using CalculusWithJuliaSquared
import IntervalArithmetic # rigorous enclosures, for the squeeze theorem
plotly()16 Limits, issues, extensions of the concept
This section uses the following add-on packages:
CalculusWithJuliaSquared re-exports Plots and Symbolics, so neither needs a using of its own. Limits here are taken with symlim, introduced in the previous chapter: it returns the value and the route it took, so each answer says how it was reached.
The limit of a function at \(c\) need not exist for one of many different reasons. Some of these reasons can be handled with extensions to the concept of the limit, others are just problematic in terms of limits. This section covers examples of each.
Let’s begin with a function that is just problematic. Consider
\[ f(x) = \sin(\frac{1}{x}) \]
As this is a composition of nice functions it will have a limit everywhere except possibly when \(x=0\), as then \(1/x\) may not have a limit. So rather than talk about where it is nice, let’s consider the question of whether a limit exists at \(c=0\).
Figure 16.1 shows the issue:
The graph oscillates between \(-1\) and \(1\) infinitely many times on this interval - so many times, that no matter how close one zooms in, the graph on the screen will fail to capture them all. Graphically, there is no single value of \(L\) that the function gets close to, as it varies between all the values in \([-1,1]\) as \(x\) gets close to \(0\). A simple proof that there is no limit, is to take any \(\epsilon\) less than \(1\), then with any \(\delta > 0\), there are infinitely many \(x\) values where \(f(x)=1\) and infinitely many where \(f(x) = -1\). That is, there is no \(L\) with \(|f(x) - L| < \epsilon\) when \(\epsilon\) is less than \(1\) for all \(x\) near \(0\).
This function basically has too many values it gets close to. Another favorite example of such a function is the function that is \(0\) if \(x\) is rational and \(1\) if not. This function will have no limit anywhere, not just at \(0\), and for basically the same reason as above.
The issue isn’t oscillation though. Take, for example, the function \(f(x) = x \cdot \sin(1/x)\). This function again has a limit everywhere save possibly \(0\). But in this case, there is a limit at \(0\) of \(0\). This is because, the following is true:
\[ -|x| \leq x \sin(1/x) \leq |x|. \]
The following figure illustrates:
f(x) = x * sin(1/x)
plot(f, -1, 1; label="f")
plot!(abs; label="|.|")
plot!(x -> -abs(x); label="-|.|")The squeeze theorem of calculus is the formal reason \(f\) has a limit at \(0\), as both the upper function, \(|x|\), and the lower function, \(-|x|\), have a limit of \(0\) at \(0\).
16.1 Right and left limits
Another example where \(f(x)\) has no limit is the function \(f(x) = x /|x|, x \neq 0\). This function is \(-1\) for negative \(x\) and \(1\) for positive \(x\). Again, this function will have a limit everywhere except possibly at \(x=0\), where division by \(0\) is possible.
It’s graph is
f(x) = abs(x)/x
plot(f, -2, 2)The sharp jump at \(0\) is misleading - again, the plotting algorithm just connects the points, it doesn’t handle what is a fundamental discontinuity well - the function is not defined at \(0\) and jumps from \(-1\) to \(1\) there. Similarly to our example of \(\sin(1/x)\), near \(0\) the function get’s close to both \(1\) and \(-1\), so will have no limit. (Again, just take \(\epsilon\) smaller than \(1\).)
But unlike the previous example, this function would have a limit if the definition didn’t consider values of \(x\) on both sides of \(c\). The limit on the right side would be \(1\), the limit on the left side would be \(-1\). This distinction is useful, so there is an extension of the idea of a limit to one-sided limits.
Let’s loosen up the language in the definition of a limit to read:
The \(\epsilon-\delta\) definition has \(V = (L-\epsilon, L + \epsilon)\) and \(U=(c-\delta, c+\delta)\). This is a rewriting of \(L-\epsilon < f(x) < L + \epsilon\) as \(|f(x) - L| < \epsilon\).
Now for the definition:
symlim takes a side keyword — :right or :left — to request a one-sided limit. The default, side = :both, is the two-sided limit of the definition above.
@variables x1-element Vector{Num}:
xf(x) = abs(x)/x
symlim(f(x), x, 0; side = :right), symlim(f(x), x, 0; side = :left)((1//1, :series), ((-1//1), :series))
The two one-sided limits are \(1\) and \(-1\), exactly as the graph suggests. Both come back by the :series route: approached from one side at a time, \(|x|\) is just \(x\) or \(-x\), and \(x/x\) and \(-x/x\) are constants.
Ask instead for the two-sided limit and the answer is a refusal:
f(x) = abs(x)/x
symlim(f(x), x, 0)(nothing, :sides_disagree)The relation between the two concepts is that a function has a limit at \(c\) if and only if the left and right limits exist and are equal. This function \(f\) has both existing, but the two limits are not equal.
There are other such functions that jump. Another useful one is the floor function, which just rounds down to the nearest integer. A graph shows the basic shape:
plot(floor, -5,5)Again, the (nearly) vertical lines are an artifact of the graphing algorithm and not actual points that solve \(y=f(x)\). The floor function has limits except at the integers. There the left and right limits differ.
Consider the limit at \(c=0\). If \(0 < x < 1/2\), say, then \(f(x) = 0\) as we round down, so the right limit will be \(0\). However, if \(-1/2 < x < 0\), then the \(f(x) = -1\), again as we round down, so the left limit will be \(-1\). Again, with this example both the left and right limits exists, but at the integer values they are not equal, as they differ by 1.
Some functions only have one-sided limits as they are not defined in an interval around \(c\). There are many examples, but we will take \(f(x) = x^x\) and consider \(c=0\). This function is not well defined for all \(x < 0\), so it is typical to just take the domain to be \(x > 0\). Still it has a right limit \(\lim_{x \rightarrow 0+} x^x = 1\), which we can verify:
symlim(x^x, x, 0; side = :right)(1, :substitution)This agrees with the IEEE convention of assigning 0^0 to be 1.
However, not all such functions with indeterminate forms of \(0^0\) will have a limit of \(1\).
Example
Consider this funny graph:
Describe the limits at \(-1\), \(0\), and \(1\).
At \(-1\) we see a jump, there is no limit but instead a left limit of 1 and a right limit appearing to be \(1/2\).
At \(0\) we see a limit of \(1\).
Finally, at \(1\) again there is a jump, so no limit. Instead the left limit is about \(-1\) and the right limit \(1\).
16.2 Limits at infinity
The loose definition of a horizontal asymptote is “a line such that the distance between the curve and the line approaches \(0\) as they tend to infinity.” This sounds like it should be defined by a limit. The issue is, that the limit would be at \(\pm\infty\) and not some finite \(c\). This requires the idea of a neighborhood of \(c\), \(0 < |x-c| < \delta\), to be reworked.
The basic idea for a limit at \(+\infty\) is that for any \(\epsilon\), there exists an \(M\) such that when \(x > M\) it must be that \(|f(x) - L| < \epsilon\). For a horizontal asymptote, the line would be \(y=L\). Similarly a limit at \(-\infty\) can be defined with \(x < M\) being the condition.
Let’s consider some cases.
The function \(f(x) = \sin(x)\) will not have a limit at \(+\infty\) for exactly the same reason that \(f(x) = \sin(1/x)\) does not have a limit at \(c=0\) - it just oscillates between \(-1\) and \(1\) so never eventually gets close to a single value.
symlim declines, which is the correct answer — there is no limit to report:
symlim(sin(x), x, Inf)(nothing, :unresolved)(We use Julia’s Inf for \(\infty\). symlim takes it directly as the limit point.)
The refusal is earned rather than assumed, and the previous chapter showed how: interval arithmetic encloses \(\sin\) over a whole neighbourhood at once, and here the enclosure is \([-1, 1]\) no matter how far out we start. An enclosure that never collapses is a function with no limit:
for M in (10, 100, 1000)
box = IntervalArithmetic.interval(float(M), float(10M))
println(rpad("x ∈ [$M, $(10M)]", 20), sin(box))
endx ∈ [10, 100] [-1.0, 1.0]_com
x ∈ [100, 1000] [-1.0, 1.0]_com
x ∈ [1000, 10000] [-1.0, 1.0]_comWhere SymPy reports this as an AccumBounds(-1, 1) object, the interval \([-1,1]\) above is the same statement, arrived at by computation rather than by a special return type.
However, a damped oscillation, such as \(f(x) = e^{-x} \sin(x)\) will have a limit:
symlim(exp(-x)*sin(x), x, Inf)(0.0, :squeeze)The :squeeze route names the reason: \(-e^{-x} \leq e^{-x}\sin(x) \leq e^{-x}\), and both bounds go to \(0\). The oscillation never stops, but the room it has to oscillate in does.
We have rational functions will have the expected limit. In this example \(m = n\), so we get a horizontal asymptote that is not \(y=0\):
symlim((x^2 - 2x + 2)/(4x^2 + 3x - 2), x, Inf)(1//4, :reciprocal)The answer is the exact \(1/4\), not \(0.25\). That is the :reciprocal route at work: for a ratio of polynomials at infinity, substituting \(x \to 1/u\) turns the limit into one at \(0\), where the leading coefficients divide exactly.
Though rational functions can have only one (at most) horizontal asymptote, this isn’t true for all functions. Consider the following \(f(x) = x / \sqrt{x^2 + 4}\). It has different limits depending if \(x\) goes to \(\infty\) or negative \(\infty\).
Handed over as written, this one is refused:
symlim(x / sqrt(x^2 + 4), x, Inf)(nothing, :unresolved)The square root puts it outside every exact route: it is not a ratio of polynomials, so :reciprocal does not apply, and the Gruntz engine is built for log-exponential forms. The fix is the reduction this section just described — trade the limit at \(\infty\) for a limit at \(0\) by setting \(u = 1/x\). Dividing through by \(x\) (which is \(+\sqrt{x^2}\) for \(x>0\)),
\[ \frac{x}{\sqrt{x^2+4}} = \frac{1}{\sqrt{1 + 4/x^2}} = \frac{1}{\sqrt{1 + 4u^2}}, \]
and as \(x \rightarrow \infty\) we have \(u \rightarrow 0\):
@variables u
symlim(1/sqrt(1 + 4u^2), u, 0)(1, :substitution)For \(x \rightarrow -\infty\) the only change is that \(\sqrt{x^2} = -x\), which flips the sign:
@variables u
symlim(-1/sqrt(1 + 4u^2), u, 0)(-1, :substitution)Both are exact, and the work of getting there is the mathematics — recognising that \(\sqrt{x^2}\) is \(|x|\), and that \(|x|\) has a sign that depends on which infinity you are heading toward. That sign is the reason the two limits differ.
(A simpler example showing this behavior is just the function \(x/|x|\) considered earlier.)
Example: Limits at infinity and right limits at \(0\)
Given a function \(f\) the question of whether this exists:
\[ \lim_{x \rightarrow \infty} f(x) \]
can be reduced to the question of whether this limit exists:
\[ \lim_{x \rightarrow 0+} f(1/x) \]
So whether \(\lim_{x \rightarrow 0+} \sin(1/x)\) exists is equivalent to whether \(\lim_{x\rightarrow \infty} \sin(x)\) exists, which clearly does not due to the oscillatory nature of \(\sin(x)\).
Similarly, one can make this reduction
\[ \lim_{x \rightarrow c+} f(x) = \lim_{x \rightarrow 0+} f(c + x) = \lim_{x \rightarrow \infty} f(c + \frac{1}{x}). \]
That is, right limits can be analyzed as limits at \(\infty\) or right limits at \(0\), should that prove more convenient.
16.3 Limits of infinity
Vertical asymptotes are nicely defined with, as with horizontal asymptotes, by the graph getting close to some line. However, the formal definition of a limit won’t be the same. For a vertical asymptote, the value of \(f(x)\) heads towards positive or negative infinity, not some finite \(L\). As such, a neighborhood like \((L-\epsilon, L+\epsilon)\) will no longer make sense, rather we replace it with an expression like \((M, \infty)\) or \((-\infty, M)\). As in: the limit of \(f(x)\) as \(x\) approaches \(c\) is infinity if for every \(M > 0\) there exists a \(\delta>0\) such that if \(0 < |x-c| < \delta\) then \(f(x) > M\). Approaching \(-\infty\) would conclude with \(f(x) < -M\) for \(M>0\).
Examples
Consider the function \(f(x) = 1/x^2\). This will have a limit at every point except possibly \(0\), where division by \(0\) is possible. In this case, there is a vertical asymptote, as seen in the following graph. The limit at \(0\) is \(\infty\), in the extended sense above. For \(M>0\), we can take any \(0 < \delta < 1/\sqrt{M}\). The following graph shows \(M=25\) where the function values are outside of the box, as \(f(x) > M\) for those \(x\) values with \(0 < |x-0| < 1/\sqrt{M}\).
The function \(f(x)=1/x\) requires us to talk about left and right limits of infinity, with the natural generalization. We can see that the left limit at \(0\) is \(-\infty\) and the right limit \(\infty\):
And symlim agrees:
f(x) = 1/x
symlim(f(x), x, 0; side = :left), symlim(f(x), x, 0; side = :right)((-Inf, :gruntz), (Inf, :gruntz))Consider the function \(g(x) = x^x(1 + \log(x)), x > 0\). Does this have a right limit at \(0\)?
A quick graph shows that a limit may be \(-\infty\):
g(x) = x^x * (1 + log(x))
plot(g, 1/100, 1)We can check:
symlim(g(x), x, 0; side = :right)(-Inf, :divergent_numeric)Here the route is :divergent_numeric: no series or engine settled it, so the answer rests on the values themselves growing without bound along the side approached. That is a weaker kind of evidence than an exact cancellation, and naming it says so.
16.4 Limits of sequences
After all this, we still can’t formalize the basic question asked in the introduction to limits: what is the area contained in a parabola. For that we developed a sequence of sums: \(s_n = 1/2 \cdot((1/4)^0 + (1/4)^1 + (1/4)^2 + \cdots + (1/4)^n)\). This isn’t a function of real \(x\), but rather depends only on non-negative integer values of \(n\). However, the same idea as a limit at infinity can be used to define a limit.
Let \(a_0,a_1, a_2, \dots, a_n, \dots\) be a sequence of values indexed by \(n\). We have \(\lim_{n \rightarrow \infty} a_n = L\) if for every \(\epsilon > 0\) there exists an \(M>0\) where if \(n > M\) then \(|a_n - L| < \epsilon\).
Common language is the sequence converges when the limit exists and otherwise diverges.
The above is essentially the same as a limit at infinity for a function, but in this case the function’s domain is only the non-negative integers.
Limits of sequences are computed the same way, once the sequence is written as an expression in \(n\). There is a wrinkle: Symbolics has no symbolic summation — no way to write \(\sum_{i=0}^{n}\) with a symbolic upper bound. Julia’s own sum works, but only over a range that is known:
@variables x
sum(x^i for i in 0:3) # a concrete range: fine
\[ 1 + x + x^{2} + x^{3} \]
Writing 0:n for a symbolic n fails before Symbolics is even involved, since the range itself cannot be built. So we supply the closed form ourselves — which for this sum is the geometric series formula the chapter on limits already derived:
\[ s_n = \frac{1}{2}\sum_{i=0}^{n} \left(\frac{1}{4}\right)^i = \frac{1}{2}\cdot\frac{1 - (1/4)^{n+1}}{1 - 1/4} = \frac{2}{3}\left(1 - \left(\tfrac{1}{4}\right)^{n+1}\right). \]
Now the limit is a limit of an ordinary expression in \(n\). Written with the power as it stands, though, the exponent is symbolic, and a symbolic exponent blocks every exact route:
@variables n
symlim((2//3) * (1 - (1//4)^(n+1)), n, Inf)(0.6666666666666666, :squeeze)The value is right, and the route says how it was reached: :squeeze bounds the expression with interval arithmetic and reports the midpoint of a box that has closed on \(2/3\) — a float, because a float is all a bound knows. Nothing here was derived. Rewriting \(a^m\) as \(e^{m\log a}\) — the same move used twice in the previous chapter — puts the expression in log-exponential form, which is the domain the Gruntz engine is defined on. That rewrite is scaffolding for the engine, not mathematics (the two forms are the same function); what changes mathematically is that the exact route now applies, and the answer comes back derived:
@variables n
symlim((2//3) * (1 - exp((n+1) * log(1//4))), n, Inf)(2//3, :gruntz)\(2/3\), the area under the parabola, as an exact rational.
Example
The limit
\[ \lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1, \]
is an important limit. Using the definition of \(e^x\) by an infinite sequence:
\[ e^x = \lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n, \]
we can establish the limit using the squeeze theorem. First,
\[ A = |(1 + \frac{x}{n})^n - 1 - x| = |\Sigma_{k=0}^n {n \choose k}(\frac{x}{n})^k - 1 - x| = |\Sigma_{k=2}^n {n \choose k}(\frac{x}{n})^k|, \]
the first two sums cancelling off. The above comes from the binomial expansion theorem for a polynomial. Now \({n \choose k} \leq n^k\)so we have
\[ A \leq \Sigma_{k=2}^n |x|^k = |x|^2 \frac{1 - |x|^{n+1}}{1 - |x|} \leq \frac{|x|^2}{1 - |x|}. \]
using the geometric sum formula with \(x \approx 0\) (and not \(1\)):
\[ \sum_{i=0}^{n} x^i = \frac{1 - x^{n+1}}{1 - x}. \]
As above, the symbolic upper bound puts this beyond sum, but a concrete range shows the identity holding:
@variables x
n = 4
lhs = sum(x^i for i in 0:n)
rhs = simplify((1 - x^(n+1)) / (1 - x))
lhs, rhs, isequal(expand(lhs), expand(rhs))(1 + x + x^2 + x^3 + x^4, 1 + x + x^2 + x^3 + x^4, true)
As this holds for all \(n\), as \(n\) goes to \(\infty\) we have:
\[ |e^x - 1 - x| \leq \frac{|x|^2}{1 - |x|} \]
Dividing both sides by \(x\) and noting that as \(x \rightarrow 0\), \(|x|/(1-|x|)\) goes to \(0\) by continuity, the squeeze theorem gives the limit:
\[ \lim_{x \rightarrow 0} \frac{e^x -1}{x} - 1 = 0. \]
That \({n \choose k} \leq n^k\) can be viewed as the left side counts the number of combinations of \(k\) choices from \(n\) distinct items, which is less than the number of permutations of \(k\) choices, which is less than the number of choices of \(k\) items from \(n\) distinct ones without replacement – what \(n^k\) counts.
16.5 Summary
The following table captures the various changes to the definition of the limit to accommodate some of the possible behaviors.
| Type | Notation | V | U |
|---|---|---|---|
| limit | \(\lim_{x\rightarrow c}f(x) = L\) | \((L-\epsilon, L+\epsilon)\) | \((c - \delta, c+\delta)\) |
| right limit | \(\lim_{x\rightarrow c+}f(x) = L\) | \((L-\epsilon, L+\epsilon)\) | \((c, c+\delta)\) |
| left limit | \(\lim_{x\rightarrow c-}f(x) = L\) | \((L-\epsilon, L+\epsilon)\) | \((c - \delta, c)\) |
| limit at \(\infty\) | \(\lim_{x\rightarrow \infty}f(x) = L\) | \((L-\epsilon, L+\epsilon)\) | \((M, \infty)\) |
| limit at \(-\infty\) | \(\lim_{x\rightarrow -\infty}f(x) = L\) | \((L-\epsilon, L+\epsilon)\) | \((-\infty, M)\) |
| limit of \(\infty\) | \(\lim_{x\rightarrow c}f(x) = \infty\) | \((M, \infty)\) | \((c - \delta, c+\delta)\) |
| limit of \(-\infty\) | \(\lim_{x\rightarrow c}f(x) = -\infty\) | \((-\infty, M)\) | \((c - \delta, c+\delta)\) |
| limit of a sequence | \(\lim_{n \rightarrow \infty} a_n = L\) | \((L-\epsilon, L+\epsilon)\) | \((M, \infty)\) |
Ross summarizes this by enumerating the 15 different related definitions for \(\lim_{x \rightarrow a} f(x) = L\) that arise from \(L\) being either finite, \(-\infty\), or \(+\infty\) and \(a\) being any of \(c\), \(c-\), \(c+\), \(-\infty\), or \(+\infty\).
16.6 Rates of growth
Consider two functions \(f\) and \(g\) to be comparable if there are positive integers \(m\) and \(n\) with both
\[ \lim_{x \rightarrow \infty} \frac{f(x)^m}{g(x)} = \infty \quad\text{and } \lim_{x \rightarrow \infty} \frac{g(x)^n}{f(x)} = \infty. \]
The first says \(g\) is eventually bounded by a power of \(f\), the second that \(f\) is eventually bounded by a power of \(g\).
Here we consider which families of functions are comparable.
First consider \(f(x) = x^3\) and \(g(x) = x^4\). We can take \(m=2\) and \(n=1\) to verify \(f\) and \(g\) are comparable:
@variables x
fx, gx = x^3, x^4
symlim(fx^2/gx, x, Inf), symlim(gx^1/fx, x, Inf)((Inf, :reciprocal), (Inf, :reciprocal))Both are ratios of polynomials, so both go by :reciprocal — the exact route from earlier in the chapter.
Similarly for any pairs of powers, so we could conclude \(f(x) = x^n\) and \(g(x) =x^m\) are comparable. (However, as is easily observed, for \(m\) and \(n\) both positive integers \(\lim_{x \rightarrow \infty} x^{m+n}/x^m = \infty\) and \(\lim_{x \rightarrow \infty} x^{m}/x^{m+n} = 0\), consistent with our discussion on rational functions that higher-order polynomials dominate lower-order polynomials.)
Now consider \(f(x) = x\) and \(g(x) = \log(x)\). These are not compatible as there will be no \(n\) large enough. We might say \(x\) dominates \(\log(x)\).
We want this for every \(n\), so \(n\) must stay symbolic. A symbolic exponent blocks every exact route — symlim declines rather than guess, since the order of \(x^k\) is \(k\) itself. But \(\log(x)^n/x\) can be rewritten so that \(n\) is no longer an exponent. Put \(u = \log(x)\), so that \(x = e^u\) and \(u \rightarrow \infty\) with \(x\):
\[ \frac{\log(x)^n}{x} = \frac{u^n}{e^u} = e^{\,n\log(u) - u}. \]
Now \(n\) is a coefficient inside an exponential, which is precisely the log-exponential form the Gruntz engine handles:
@variables u n
symlim(exp(n*log(u) - u), u, Inf)(0, :gruntz)Zero, for every \(n\) at once, with no assumption placed on \(n\) at all. As \(x\) could be replaced by any monomial \(x^k\), we can say “powers” grow faster than “logarithms”.
Now consider \(f(x)=x\) and \(g(x) = e^x\). These are not compatible as there will be no \(m\) large enough:
The same rewrite applies, and here it is even more direct: \(x^m/e^x = e^{m\log(x) - x}\).
@variables x m
symlim(exp(m*log(x) - x), x, Inf)(0, :gruntz)That is \(e^x\) grows faster than any power of \(x\).
Now, if \(a, b > 1\) then \(f(x) = a^x\) and \(g(x) = b^x\) will be comparable. Take \(m\) so that \(a^m > b\) and \(n\) so that \(b^n > a\) as then, say,
\[ \frac{(a^x)^m}{b^x} = \frac{a^{xm}}{b^x} = \frac{(a^m)^x}{b^x} = (\frac{a^m}{b})^x, \]
which will go to \(\infty\) as \(x \rightarrow \infty\) as \(a^m/b > 1\).
Finally, consider \(f(x) = \exp(x^2)\) and \(g(x) = \exp(x)^2\). Are these comparable? No, as no \(n\) is large enough:
Once more the exponents collapse into the exponential: \((e^x)^{2n}/e^{x^2} = e^{2nx - x^2}\).
@variables x n
symlim(exp(2n*x - x^2), x, Inf)(0, :gruntz)A negative test for compatibility is the following: if
\[ \lim_{x \rightarrow \infty} \frac{\log(|f(x)|)}{\log(|g(x)|)} = 0, \]
Then \(f\) and \(g\) are not compatible (and \(g\) grows faster than \(f\)). Applying this to the last two values of \(f\) and \(g\), we have
\[ \lim_{x \rightarrow \infty}\frac{\log(\exp(x)^2)}{\log(\exp(x^2))} = \lim_{x \rightarrow \infty}\frac{2\log(\exp(x))}{x^2} = \lim_{x \rightarrow \infty}\frac{2x}{x^2} = 0, \]
so \(f(x) = \exp(x^2)\) grows faster than \(g(x) = \exp(x)^2\).
Keeping in mind that logarithms grow slower than powers which grow slower than exponentials (\(a > 1\)) can help understand growth at \(\infty\) as a comparison of leading terms does for rational functions.
We can immediately put this to use to compute \(\lim_{x\rightarrow 0+} x^x\). We first express this problem using \(x^x = (\exp(\ln(x)))^x = e^{x\ln(x)}\). Rewriting \(u(x) = \exp(\ln(u(x)))\), which only uses the basic inverse relation between the two functions, can often be a useful step.
As \(f(x) = e^x\) is a suitably nice function (continuous) so that the limit of a composition can be computed through the limit of the inside function, \(x\ln(x)\), it is enough to see what \(\lim_{x\rightarrow 0+} x\ln(x)\) is. We re-express this as a limit at \(\infty\)
\[ \lim_{x\rightarrow 0+} x\ln(x) = \lim_{x \rightarrow \infty} (1/x)\ln(1/x) = \lim_{x \rightarrow \infty} \frac{-\ln(x)}{x} = 0 \]
The last equality follows, as the function \(x\) dominates the function \(\ln(x)\). So by the limit rule involving compositions we have: \(\lim_{x\rightarrow 0+} x^x = e^0 = 1\).
16.7 Questions
Question
Select the graph for which the limit at \(a\) is infinite.
Question
Select the graph for which the limit at \(\infty\) appears to be defined.
Question
Consider the function \(f(x) = \sqrt{x}\).
Does this function have a limit at every \(c > 0\)?
Does this function have a limit at \(c=0\)?
Does this function have a right limit at \(c=0\)?
Does this function have a left limit at \(c=0\)?
Question
Find \(\lim_{x \rightarrow \infty} \sin(x)/x\).
Question
Find \(\lim_{x \rightarrow \infty} (1-\cos(x))/x^2\).
Question
Find \(\lim_{x \rightarrow \infty} \log(x)/x\).
Question
Find \(\lim_{x \rightarrow 2+} (x-3)/(x-2)\).
Find \(\lim_{x \rightarrow -3-} (x-3)/(x+3)\).
Question
Let \(f(x) = \exp(x + \exp(-x^2))\) and \(g(x) = \exp(-x^2)\). Compute:
\[ \lim_{x \rightarrow \infty} \frac{\ln(f(x))}{\ln(g(x))}. \]
Question
Consider the following expression:
@variables x
1/(exp(-x + exp(-x))) - exp(x)
\[ - e^{x} + \frac{1}{e^{e^{ - x} - x}} \]
We want to find the limit, \(L\), as \(x \rightarrow \infty\), which we assume exists below.
We first rewrite the expression using \(w\) for \(e^{-x}\). Since \(1/e^{-x + e^{-x}} = e^{x}e^{-e^{-x}}\) and \(e^{x} = 1/w\), the whole expression becomes
\[ e^{x}\left(e^{-w} - 1\right) = \frac{e^{-w} - 1}{w}. \]
As \(x \rightarrow \infty\), \(w \rightarrow 0+\), so the limit at \(0\) of this is of interest:
@variables w
symlim((exp(-w) - 1)/w, w, 0)(-1//1, :series)Exactly \(-1\), by the series route. The awkward-looking manoeuvre in the original — subtract the linear polynomial you expect and check the remainder vanishes — works too, and is worth seeing because it generalises:
@variables w
symlim((exp(-w) - 1)/w - (w/2 - 1), w, 0)(0, :series)The remainder goes to \(0\), so \((e^{-w}-1)/w\) and \(w/2 - 1\) have the same limit, and the latter is read off by inspection at \(w = 0\).
\(L\) is:
(This awkward approach is generalizable: replacing the limit as \(w \rightarrow 0\) of an expression with the limit of a polynomial in w that is easy to identify.)
Question
Some limits depend on the specific value of a parameter, and cannot be answered while it stays symbolic. Consider
\[ \lim_{x \rightarrow 0+} \frac{\sin(\sin(x^2))}{x^k}. \]
With a symbolic k, symlim declines:
@variables x k
symlim(sin(sin(x^2))/x^k, x, 0; side = :right)(nothing, :unresolved)This is a refusal on principle, not a shortcoming. The order of \(x^k\) is \(k\), so ranking the numerator’s leading order against the denominator’s is meaningless until \(k\) is known — and the answer genuinely differs for different \(k\). Unlike the growth-rate limits earlier in this section, no rewriting can help, because here the exponent really is the whole question.
Give it a value and it answers at once. Try \(k = 1, 2, 3\):
@variables x
[symlim(sin(sin(x^2))/x^k, x, 0; side = :right) for k in 1:3]3-element Vector{Tuple{Real, Symbol}}:
(0, :series)
(1//1, :series)
(Inf, :divergent_numeric)For which value(s) of \(k\) in \(1,2,3\) is the limit \(0\)?
Question: No limit
Some functions do not have a limit. Make a graph of \(\sin(1/x)\) from \(0.0001\) to \(1\) and look at the output. Why does a limit not exist?
Question \(0^0\) is not always \(1\)
Is the form \(0^0\) really indeterminate? As mentioned 0^0 evaluates to 1.
Consider this limit:
\[ \lim_{x \rightarrow 0+} x^{k\cdot x} = L. \]
Consider different values of \(k\) to see if this limit depends on \(k\) or not. What is \(L\)?
Now, consider this limit:
\[ \lim_{x \rightarrow 0+} x^{1/\log_k(x)} = L. \]
In julia, \(\log_k(x)\) is found with log(k,x). The default, log(x) takes \(k=e\) so gives the natural log. So, we would define h, for a given k, with
k = 10 # say. Replace with actual value
h(x) = x^(1/log(k, x))h (generic function with 1 method)Consider different values of \(k\) to see if the limit depends on \(k\) or not. What is \(L\)?
Question
Limits of infinity at infinity. We could define this concept quite easily mashing together the two definitions. Suppose we did. Which of these ratios would have a limit of infinity at infinity:
\[ x^4/x^3,\quad x^{100+1}/x^{100}, \quad x/\log(x), \quad 3^x / 2^x, \quad e^x/x^{100} \]
Question
A slant asymptote is a line \(mx + b\) for which the graph of \(f(x)\) gets close to as \(x\) gets large. We can’t express this directly as a limit, as “\(L\)” is not a number. How might we?
Question
Suppose a sequence of points \(x_n\) converges to \(a\) in the limiting sense. For a function \(f(x)\), the sequence of points \(f(x_n)\) may or may not converge. One alternative definition of a limit due to Heine is that \(\lim_{x \rightarrow a}f(x) = L\) if and only if all sequences \(x_n \rightarrow a\) have \(f(x_n) \rightarrow L\).
Consider the function \(f(x) = \sin(1/x)\), \(a=0\), and the two sequences implicitly defined by \(1/x_n = \pi/2 + n \cdot (2\pi)\) and \(1/y_n = 3\pi/2 + n \cdot(2\pi)\), \(n = 0, 1, 2, \dots\).
What is \(\lim_{x_n \rightarrow 0} f(x_n)\)?
What is \(\lim_{y_n \rightarrow 0} f(y_n)\)?
This shows that