# How does the chain rule for limits work?

I have to evaluate the limit of this function,

$$\lim_{x\to0^+} \arctan(\ln x)$$

I already know the answer, it's $-\dfrac{π}{2}$, but the only part I don't get it, how does it come to that? I did the following steps:

$$\lim_{x\to0^+} \arctan(\ln x) = \arctan\left(\lim_{x\to0^+} \ln x\right)$$

The limit of $\ln(x)$ when $x$ approaces $0^+$ is negative infinity, wouldn't that mean the answer we're looking for is arctan of negative infinity, which is something we can't find?

Still, it goes to:

$$\lim_{x\to -\infty} \arctan (x) = -\dfrac{\pi}{2}$$, which is how the answer seem to work? How does this happen? And, how does the chain rule come in all this?

• You don't mean "chain rule" here, rather "composition of functions".
– Doc
Sep 24, 2013 at 1:26
• @Doc : Any rule about how to apply an operation to a composite of functions may be called a chain rule. The chain rule for differentiation is most famous, but there's also a chain rule for limits. (Similarly for product rules, sum rules, etc.) Feb 16, 2019 at 5:12
• Where you read that? @TobyBartels Jun 14, 2020 at 22:47
• @EduardoS. : You can read it at google.com/search?q=%22chain+rule+for+limits%22 (or google.com/search?tbm=bks&q=%22chain+rule+for+limits%22 if you only want to see it in textbooks). The book search gives very few hits, but it probably leaves out most textbooks that use the term; a book that lists a ‘chain rule’ in a chapter on ‘limits’ without using the exact phrase ‘chain rule for limits’ is harder to search for. (I can probably find such a book when I get back to my office in August.) Jun 15, 2020 at 21:52
• There are some more results at google.com/search?q=%22limit+chain+rule%22 (but the results at google.com/search?tbm=bks&q=%22limit+chain+rule%22 are all false positives, so ‘limit chain rule’ seems to be only an informal term). Jun 15, 2020 at 22:07

Look at the expression $$\lim_{x \to 0^+} \arctan(\ln x)$$ Let $u = \ln x$. Then $u \to -\infty$ as $x \to 0^+$. So we can substitute $u$ for $\ln x$ and $u \to -\infty$ for $x \to 0^+$ to obtain $$\lim_{x \to 0^+} \arctan(\ln x) = \lim_{u \to -\infty} \arctan(u)$$ This evaluates to $-\dfrac{\pi}{2}$.

All we did was substitute a new variable; nothing too in-depth!

• Thank you, that makes sense! Sep 24, 2013 at 1:38
• Unfortunately, this sort of substitution does not in general work with limits. It works here because the intermediate limit is infinite (and the logarithm function is finite as $x \to 0^+$); it also works with a finite intermediate limit if the inner function never takes the value of that limit (or if either function is continuous, of course); probably there are other situations where it's valid. But it fails if the inner function is constant and the outer function is discontinuous at that value! Jan 29, 2014 at 23:03
• Here is a fancier situation where it fails: $$lim_{x \to 0} [-|x \sin(1/x)|].$$ (Here $[t]$ is the floor of $t$, the largest integer not larger than $t$.) As $x \to 0$, $x \sin(1/x) \to 0$ (a classic example of the squeeze/sandwich theorem). As $u \to 0$, $[-|u|] \to -1$ (because $-|u| < 0$ for $u \ne 0$). But as $x \to 0$, $[-|x sin(1/x)|]$ has no limit, because it takes the value $0$ at $x = 1/\pi, 1/(2\pi), 1/(3\pi), \ldots$, which can be arbitrarily close to $0$. Jan 29, 2014 at 23:03
• I can no longer edit my first comment above, but it has a minor error; where it says ‘or if either function is continuous’, it should say ‘or if the outer function is continuous’. Jan 29, 2014 at 23:53

As $x$ approaches $0$ from the right, $\ln x$ becomes very large negative. As $w$ becomes very large negative, $\arctan w$ approaches $-\frac{\pi}{2}$.

• My question is this, how does arctan(−∞) suddenly become limx→−∞arctan(w) ? Sep 24, 2013 at 1:29
• you cant take the arctan of -infinity, -infinity is not a number, but you can take a limit as it approaches infinity
– YYC
Sep 24, 2013 at 1:30

As $x$ approaches $-\dfrac{π}{2}$ from the right it will approach negative infinity, so $$\lim_{x\to-\infty}\arctan (x)=-\dfrac{π}{2}$$

The theorem says that if $\lim_{x \to c} g(x) = w$ and $f$ is continous at $w$, then $\lim_{x \to c} f(g(x)) = f( \lim_{x \to c} g(x))$. Clearly, the statement "$f$ is continous at $w$ assumes that $w$ is a real number. This is the premise your limit fails. Your $w$ is "$\infty"$.

I see that some years ago, I commented on the accepted answer (by @Clive Newstead) and said why it was incomplete. And there is an answer by @Ovi explaining why a common chain-rule theorem for limits doesn't apply. But no answer yet states a precise, true theorem about limits that does apply here to explain the answer. So let me fix that.

As Ovi noted, one theorem is that $$\lim \limits _ { x \to c } f ( g ( x ) ) = f ( w )$$ if $$w = \lim \limits _ { x \to c } g ( x )$$ exists and $$f$$ is continuous there. This is completely inapplicable, since $$w$$ is infinite in this case, and so there's no way that $$f$$ could be continuous there; we can't even say that $$f$$ is defined at $$w$$.

But another theorem is that $$\lim \limits _ { x \to c } f ( g ( x ) ) = \lim \limits _ { u \to w } f ( u )$$ if $$w = \lim \limits _ { x \to c } g ( x )$$ exists (possibly as a form of infinity) and $$w$$ is not in the range of $$g$$ (including when $$w$$ is a form of infinity), as long as $$\lim \limits _ { u \to w } f ( u )$$ also exists. And that's what we have here! (Actually, it's even possible for $$w$$ to be in the range of $$g$$ as long as there is a deleted neighbourhood of $$c$$ on which it is not; that is, whenever there is a $$\delta > 0$$ such that $$g ( x )$$ is never $$w$$ when $$0 < \lvert x - c \rvert < \delta$$.)

I should link to some reference where this theorem is stated, but I haven't found one, and writing out the general proof involves tedious cases. So instead, let me run through this example, showing how, given $$\epsilon > 0$$, to find $$\delta > 0$$ such that whenever $$0 < x - 0 < \delta$$, $$\lvert \arctan \ln x - ( - \pi / 2 ) \rvert < \epsilon$$, without using any knowledge about arctangents and logarithms other than the two relevant limits and the fine print about the range.

So given $$\epsilon _ 1 > 0$$, since $$\lim \limits _ { u \to - \infty } \arctan u = - \pi / 2$$, there exists $$\delta _ 1 > 0$$ such that whenever $$0 < - 1 / u < \delta _ 1$$, $$\lvert \arctan u + \pi / 2 \rvert < \epsilon _ 1$$. Now let $$\epsilon _ 2$$ be $$\delta _ 1$$; since $$\lim \limits _ { x \to 0 ^ + } \ln x = - \infty$$ and $$\epsilon _ 2 > 0$$, there exists $$\delta _ 2 > 0$$ such that whenever $$0 < x < \delta _ 2$$, $$0 \leq - 1 / \ln x < \epsilon _ 2$$. So if we start with $$\epsilon = \epsilon _ 1$$, in the end we want $$\delta = \delta _ 2$$.

Let's check that ths works! So given $$x$$ such that $$0 < x < \delta = \delta _ 2$$, we know that $$0 \leq - 1 / \ln x < \epsilon _ 2$$. Since $$- 1 / \ln x$$ cannot equal $$0$$ (in other words, $$- \infty$$ is not in the range of the natural logarithm, which is less trivial when you apply this reasoning to finite limits), in fact $$0 < - 1 / \ln x < \epsilon _ 2 = \delta _ 1$$. Let $$u$$ be $$\ln x$$; since $$0 < - 1 / u < \delta _ 1$$, $$\lvert \arctan u + \pi / 2 \rvert < \epsilon _ 1$$. In other words, $$\lvert \arctan \ln x + \pi / 2 \rvert < \epsilon$$, which is what we need.

tl;dr: $$\lim \limits _ { x \to c } f ( g ( x ) ) = \lim \limits _ { u \to w } f ( u )$$ if $$w = \lim \limits _ { x \to c } g ( x )$$ exists (possibly as a form of infinity) and $$w$$ is not in the range of $$g$$ (including when $$w$$ is a form of infinity), as long as $$\lim \limits _ { u \to w } f ( u )$$ also exists. (And it's also OK if $$w$$ is in the range of $$g$$ because $$g ( c ) = w$$ and/or because $$w = g ( x )$$ for values of $$x$$ far from $$c$$. And finally, if $$f$$ is continuous at $$w$$, as already noted, then no conditions apply to the range of $$g$$.)