We have defined and used the concept of limit, primarily in our development of the derivative. Recall that $\ds \lim_{x\to a}f(x)=L$ is true if, in a precise sense, $f(x)$ gets closer and closer to $L$ as $x$ gets closer and closer to $a$. While some limits are easy to see, others take some ingenuity; in particular, the limits that define derivatives are always difficult on their face, since in $$\lim_{\Delta x\to 0} {f(x+\Delta x)-f(x)\over \Delta x}$$ both the numerator and denominator approach zero. Typically this difficulty can be resolved when $f$ is a "nice'' function and we are trying to compute a derivative. Occasionally such limits are interesting for other reasons, and the limit of a fraction in which both numerator and denominator approach zero can be difficult to analyze. Now that we have the derivative available, there is another technique that can sometimes be helpful in such circumstances.

Before we introduce the technique, we will also expand our concept of limit, in two ways. When the limit of $f(x)$ as $x$ approaches $a$ does not exist, it may be useful to note in what way it does not exist. We have already talked about one such case: one-sided limits. Another case is when "$f$ goes to infinity''. We also will occasionally want to know what happens to $f$ when $x$ "goes to infinity''.

Example 4.7.1 What happens to $1/x$ as $x$ goes to 0? From the right, $1/x$ gets bigger and bigger, or goes to infinity. From the left it goes to negative infinity.

Example 4.7.2 What happens to the function $\ds \cos(1/x)$ as $x$ goes to infinity? It seems clear that as $x$ gets larger and larger, $1/x$ gets closer and closer to zero, so $\cos(1/x)$ should be getting closer and closer to $\cos(0)=1$.

As with ordinary limits, these concepts can be made precise. Roughly, we want $\ds \lim_{x\to a}f(x)=\infty$ to mean that we can make $f(x)$ arbitrarily large by making $x$ close enough to $a$, and $\ds \lim_{x\to \infty}f(x)=L$ should mean we can make $f(x)$ as close as we want to $L$ by making $x$ large enough. Compare this definition to the definition of limit in section 2.3, definition 2.3.2.

Definition 4.7.3 If $f$ is a function, we say that $\ds \lim_{x\to a}f(x)=\infty$ if for every $N>0$ there is a $\delta>0$ such that whenever $|x-a|< \delta$, $f(x)>N$. We can extend this in the obvious ways to define $\ds \lim_{x\to a}f(x)=-\infty$, $\ds \lim_{x\to a^-}f(x)=\pm\infty$, and $\ds \lim_{x\to a^+}f(x)=\pm\infty$.

Definition 4.7.4 (Limit at infinity) If $f$ is a function, we say that $\ds \lim_{x\to \infty}f(x)=L$ if for every $\epsilon>0$ there is an $N > 0$ so that whenever $x>N$, $|f(x)-L|< \epsilon$. We may similarly define $\ds \lim_{x\to-\infty}f(x)=L$, and using the idea of the previous definition, we may define $\ds \lim_{x\to\pm\infty}f(x)=\pm\infty$.

We include these definitions for completeness, but we will not explore them in detail. Suffice it to say that such limits behave in much the same way that ordinary limits do; in particular there are some analogs of theorem 2.3.6.

Now consider this limit: $$\lim_{x\to \pi}{x^2-\pi^2\over \sin x}.$$ As $x$ approaches $\pi$, both the numerator and denominator approach zero, so it is not obvious what, if anything, the quotient approaches. We can often compute such limits by application of the following theorem.

Theorem 4.7.5 (L'Hôpital's Rule) For "sufficiently nice'' functions $f(x)$ and $g(x)$, if $\ds\lim_{x\to a} f(x)= 0 = \lim_{x\to a} g(x)$ or both $\ds\lim_{x\to a} f(x)= \pm\infty$ and $\lim_{x\to a} g(x)=\pm\infty$, and if $\ds\lim_{x\to a}{f'(x)\over g'(x)}$ exists, then $\ds\lim_{x\to a}{f(x)\over g(x)}=\lim_{x\to a}{f'(x)\over g'(x)}$. This remains true if "$x\to a$'' is replaced by "$x\to \infty$'' or "$x\to -\infty$''.

This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here. We also will not need to worry about the precise definition of "sufficiently nice'', as the functions we encounter will be suitable.

Example 4.7.6 Compute $\ds\lim_{x\to \pi}{x^2-\pi^2\over \sin x}$ in two ways.

First we use L'Hôpital's Rule: Since the numerator and denominator both approach zero, $$\lim_{x\to \pi}{x^2-\pi^2\over \sin x}= \lim_{x\to \pi}{2x \over \cos x},$$ provided the latter exists. But in fact this is an easy limit, since the denominator now approaches $-1$, so $$\lim_{x\to \pi}{x^2-\pi^2\over \sin x}={2\pi\over -1} = -2\pi.$$

We don't really need L'Hôpital's Rule to do this limit. Rewrite it as $$\lim_{x\to \pi}(x+\pi){x-\pi\over \sin x}$$ and note that $$\lim_{x\to \pi}{x-\pi\over \sin x}= \lim_{x\to \pi}{x-\pi\over -\sin (x-\pi)}= \lim_{x\to 0}-{x\over \sin x}$$ since $x-\pi$ approaches zero as $x$ approaches $\pi$. Now $$\lim_{x\to \pi}(x+\pi){x-\pi\over \sin x}= \lim_{x\to \pi}(x+\pi)\lim_{x\to 0}-{x\over \sin x}= 2\pi(-1)=-2\pi$$ as before.

Example 4.7.7 Compute $\ds\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}$ in two ways.

As $x$ goes to infinity both the numerator and denominator go to infinity, so we may apply L'Hôpital's Rule: $$\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}= \lim_{x\to \infty}{4x-3\over 2x+47}.$$ In the second quotient, it is still the case that the numerator and denominator both go to infinity, so we are allowed to use L'Hôpital's Rule again: $$\lim_{x\to \infty}{4x-3\over 2x+47}=\lim_{x\to \infty}{4\over 2}=2.$$ So the original limit is 2 as well.

Again, we don't really need L'Hôpital's Rule, and in fact a more elementary approach is easier—we divide the numerator and denominator by $\ds x^2$: $$\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}= \lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}{{1\over x^2}\over {1\over x^2}}= \lim_{x\to \infty}{2-{3\over x}+{7\over x^2}\over 1+{47\over x}+{1\over x^2}}.$$ Now as $x$ approaches infinity, all the quotients with some power of $x$ in the denominator approach zero, leaving 2 in the numerator and 1 in the denominator, so the limit again is 2.

Example 4.7.8 Compute $\ds\lim_{x\to 0}{\sec x - 1\over \sin x}$.

Both the numerator and denominator approach zero, so applying L'Hôpital's Rule: $$\lim_{x\to 0}{\sec x - 1\over \sin x}= \lim_{x\to 0}{\sec x\tan x\over \cos x}={1\cdot 0\over 1}=0.$$

Exercises 4.7

Compute the limits.

Ex 4.7.1 $\ds\lim_{x\to 0} {\cos x -1\over \sin x}$ (answer)

Ex 4.7.2 $\ds\lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x}$ (answer)

Ex 4.7.3 $\ds\lim_{x\to0}{\sqrt{9+x}-3\over x}$ (answer)

Ex 4.7.4 $\ds\lim_{t\to1^+}{(1/t)-1\over t^2-2t+1}$ (answer)

Ex 4.7.5 $\ds\lim_{x\to2}{2-\sqrt{x+2}\over 4-x^2}$ (answer)

Ex 4.7.6 $\ds\lim_{t\to\infty}{t+5-2/t-1/t^3\over 3t+12-1/t^2}$ (answer)

Ex 4.7.7 $\ds\lim_{y\to\infty}{\sqrt{y+1}+\sqrt{y-1}\over y}$ (answer)

Ex 4.7.8 $\ds\lim_{x\to1}{\sqrt{x}-1\over \root 3\of{x}-1}$ (answer)

Ex 4.7.9 $\ds\lim_{x\to0}{(1-x)^{1/4}-1\over x}$ (answer)

Ex 4.7.10 $\ds\lim_{t\to 0}{\left(t+{1\over t}\right)((4-t)^{3/2}-8)}$ (answer)

Ex 4.7.11 $\ds\lim_{t\to 0^+}\left({1\over t}+{1\over\sqrt{t}}\right) (\sqrt{t+1}-1)$ (answer)

Ex 4.7.12 $\ds\lim_{x\to 0}{x^2\over\sqrt{2x+1}-1}$ (answer)

Ex 4.7.13 $\ds\lim_{u\to 1}{(u-1)^3\over (1/u)-u^2+3u-3}$ (answer)

Ex 4.7.14 $\ds\lim_{x\to 0}{2+(1/x)\over 3-(2/x)}$ (answer)

Ex 4.7.15 $\ds\lim_{x\to 0^+}{1+5/\sqrt{x}\over 2+1/\sqrt{x}}$ (answer)

Ex 4.7.16 $\ds\lim_{x\to 0^+}{3+x^{-1/2}+x^{-1}\over 2+4x^{-1/2}}$ (answer)

Ex 4.7.17 $\ds\lim_{x\to\infty}{x+x^{1/2}+x^{1/3}\over x^{2/3}+x^{1/4}}$ (answer)

Ex 4.7.18 $\ds\lim_{t\to\infty} {1-\sqrt{t\over t+1}\over 2-\sqrt{4t+1\over t+2}}$ (answer)

Ex 4.7.19 $\ds\lim_{t\to\infty}{1-{t\over t-1}\over 1-\sqrt{t\over t-1}}$ (answer)

Ex 4.7.20 $\ds\lim_{x\to-\infty}{x+x^{-1}\over 1+\sqrt{1-x}}$ (answer)

Ex 4.7.21 $\ds\lim_{x\to\pi/2}{\cos x\over (\pi/2)-x}$ (answer)

Ex 4.7.22 $\ds\lim_{x\to1}{x^{1/4}-1\over x}$ (answer)

Ex 4.7.23 $\ds\lim_{x\to1^+}{\sqrt{x}\over x-1}$ (answer)

Ex 4.7.24 $\ds\lim_{x\to1}{\sqrt{x}-1\over x-1}$ (answer)

Ex 4.7.25 $\ds\lim_{x\to\infty}{x^{-1}+x^{-1/2}\over x+x^{-1/2}}$ (answer)

Ex 4.7.26 $\ds\lim_{x\to\infty}{x+x^{-2}\over 2x+x^{-2}}$ (answer)

Ex 4.7.27 $\ds\lim_{x\to\infty}{5+x^{-1}\over 1+2x^{-1}}$ (answer)

Ex 4.7.28 $\ds\lim_{x\to\infty}{4x\over\sqrt{2x^2+1}}$ (answer)

Ex 4.7.29 $\ds\lim_{x\to0}{3x^2+x+2\over x-4}$ (answer)

Ex 4.7.30 $\ds\lim_{x\to0}{\sqrt{x+1}-1\over \sqrt{x+4}-2}$ (answer)

Ex 4.7.31 $\ds\lim_{x\to0}{\sqrt{x+1}-1\over \sqrt{x+2}-2}$ (answer)

Ex 4.7.32 $\ds\lim_{x\to0^+}{\sqrt{x+1}+1\over\sqrt{x+1}-1}$ (answer)

Ex 4.7.33 $\ds\lim_{x\to0}{\sqrt{x^2+1}-1\over\sqrt{x+1}-1}$ (answer)

Ex 4.7.34 $\ds\lim_{x\to\infty}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ (answer)

Ex 4.7.35 $\ds\lim_{x\to0^+}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ (answer)

Ex 4.7.36 $\ds\lim_{x\to1}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ (answer)

Ex 4.7.37 $\ds\lim_{x\to2}{x^3-6x-2\over x^3+4}$ (answer)

Ex 4.7.38 $\ds\lim_{x\to2}{x^3-6x-2\over x^3-4x}$ (answer)

Ex 4.7.39 $\ds\lim_{x\to1+}{x^3+4x+8\over 2x^3-2}$ (answer)

Ex 4.7.40 The function $\ds f(x) = {x\over\sqrt{x^2+1}}$ has two horizontal asymptotes. Find them and give a rough sketch of $f$ with its horizontal asymptotes. (answer)