Infinite limits

We can use Proposition 4.28 to get results about infinite limits.

Example 5.26   Evaluate $$\lim_{{x \to \infty}}^{}$$x log$$\left(\vphantom{1 + \frac{1}{ x}}\right.$$1 + $${\frac{{1}}{{x}}}$$$$\left.\vphantom{1 + \frac{1}{ x}}\right)$$.

Solution. 

$$\lim_{{x \to \infty}}^{} \left(\vphantom{1 + \frac{1}{ x}}\right. {\frac{{1}}{{x}}} \left.\vphantom{1 + \frac{1}{ x}}\right) \lim_{{x \to \infty}}^{} {\frac{{\log(1+y)}}{{ y}}}$$ =

$$\lim_{{y \to 0+}}^{} {\frac{{\log(1+y)}}{{ y}}}$$ =

$$\lim_{{y \to 0}}^{} {\frac{{\log(1+y)}}{{ y}}}$$ =

The last step is valid, since the final limit exists by l'Hôpital; note also that this gives another way of finding a_n = (1 + 1/n)ⁿ.

Exercise 5.27   Evaluate $$\lim_{{x \to \infty}}^{}$$x sin$$\left(\vphantom{\frac{1}{ x}}\right.$$$${\frac{{1}}{{x}}}$$$$\left.\vphantom{\frac{1}{ x}}\right)$$.

Proposition 5.28 (l'Hôpital's rule: infinite limits)   Let f and g be functions such that $\lim_{{x\to \infty}}^{}$f (x) = $\lim_{{x\to \infty}}^{}$g(x) = $\infty$, and suppose that $$\lim_{{x \to \infty}}^{}$$$${\frac{{f'(x)}}{{g'(x)}}}$$ exists. Then

$$\lim_{{x \to \infty}}^{}$$$${\frac{{f(x)}}{{g(x)}}}$$ = $$\lim_{{x \to \infty}}^{}$$$${\frac{{f'(x)}}{{g'(x)}}}$$.

Proof. (Sketch for interest -- not part of the course). Pick $\epsilon$ > 0 and choose a such that

$$\lim_{{x \to \infty}}^{}$$$$\left\vert\vphantom{\frac{f'(x)}{g'(x)} - l}\right.$$$${\frac{{f'(x)}}{{g'(x)}}}$$ - l$$\left.\vphantom{\frac{f'(x)}{g'(x)} - l}\right\vert$$ < $$\epsilon$$    for all x > a.    

Then pick K such that if x > K, then g(x) - g(a)$\ne$ 0. By Cauchy,

$${\frac{{f'(c)}}{{g'(c)}}}$$ = $${\frac{{f(x) - f(a)}}{{g(x) - g(a)}}}$$    for all x > K.    

Note that although c depends on x, we always have c > a. Then

$${\frac{{f(x)}}{{g(x)}}} {\frac{{f(x) - f(a)}}{{g(x) - g(a)}}} {\frac{{f(x)}}{{f(x) - f(a)}}} {\frac{{g(x) - g(a)}}{{g(x)}}}$$ =

$$\to \to \infty$$

$\qedsymbol$