finding the limit of $\lim_{x\to 0} \frac{e^x-1}{x^2}$

Question

$$\lim_{x\to 0} \frac{e^x-1}{x^2}$$

It is an expression in form of $\left(\frac{0}{0}\right)$. Using l'Hôpital:

$$\lim_{x\to 0} \frac{e^x}{2x}$$

The expression in form of $\left(\frac{1}{0}\right)$ so one-sided limits should be checked

$$\lim_{x\to 0^{-}} \frac{e^x}{2x}=-\infty$$

$$\lim_{x\to 0^{+}} \frac{e^x}{2x}=\infty$$

Therefore there is no limit

Is this move: The expression in form of $\left(\frac{1}{0}\right)$ so one-sided limits should be checked is right?

this Limit doesn't exist – Dr. Sonnhard Graubner Dec 08 '15 at 19:54 — Dr. Sonnhard Graubner, Dec 08 '15 at 19:54
Yes, it's a correct reasoning. – luka5z Dec 08 '15 at 19:54 — luka5z, Dec 08 '15 at 19:54

Vincenzo Oliva · Answer 1 · 2015-12-08T21:43:39.980

4

Indeed you are right. Another way to see it:

$$ \begin{align} \lim_{x\to0^{\pm}} \frac{e^x-1}{x^2} &= \lim_{x\to0^{\pm}} \frac{ \sum_{n=1}^\infty x^n/n! } {x^2}\\ &= \lim_{x\to0^{\pm}} \frac1x + \sum_{n=0}^\infty \frac{x^n}{(n+2)!}. \end{align}$$

edited Dec 08 '15 at 21:43

answered Dec 08 '15 at 20:18

Vincenzo Oliva

7,249

score 2 · Answer 2 · answered Dec 08 '15 at 20:45

2

You can note that $$ \lim_{x\to0}\frac{e^x-1}{x}=1 $$ so…

answered Dec 08 '15 at 20:45

egreg

238,574

so I can say it is $lim_{x\to 0} 1 \cdot lim_{x\to 0} \frac{1}{x}$? – gbox Dec 08 '15 at 20:53
@gbox More or less; the limit from the right is $\infty$ and the limit from the left is $-\infty$. – egreg Dec 08 '15 at 21:39

score 1 · Answer 3 · answered Dec 08 '15 at 19:55

1

This is correct. When it's $(\frac {1}{0}) $ you should check both sides of the limit, because often it won't exist.

answered Dec 08 '15 at 19:55

Will Fisher

5,112

This looks like to be a comment :-) – Math-fun Dec 08 '15 at 20:24

finding the limit of $\lim_{x\to 0} \frac{e^x-1}{x^2}$

3 Answers3