Solving $\lim_{x\to0} \frac{e^x-1-x}{x^2}$

Question

$$ = \lim_{x\to0} (\frac{e^x-1}{x} * \frac{1}{x}) -\frac{x}{x^2}$$ $$ = \lim_{x\to0} (\frac{e^x-1}{x} * \frac{1}{x}) -\frac{x}{x^2}$$ $$ = 1 * \lim_{x\to0} \frac{1}{x} -\lim_{x\to0}\frac{1}{x}$$ $$ = \infty -\infty$$

How do I actually solve this without using L'Hospital. I'm just looking for a hint, not a concrete answer.

@Salahamam_Fatima Tag: ([tag:limits-without-lhopital]) and the OP also asks that the limit be done w/o L'H. — Simply Beautiful Art, Aug 06 '17 at 23:59
Your approach has a flaw when you move from second line to third line. This is not allowed. — Paramanand Singh, Aug 07 '17 at 04:37

Simply Beautiful Art · Answer 1 · 2017-08-06T23:58:33.840

7

No series or L'Hospital's required! "DISCLAIMER": This is not a hint.

Consider $x\mapsto-x$ to get

$$L=\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^{-x}-1+x}{x^2}$$

Add these together to get

$$\begin{align}2L&=\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}\tag1\\&=\lim_{x\to0}\frac{(e^{x/2}-e^{-x/2})^2}{x^2}\tag2\\&=\left[\lim_{x\to0}\frac{e^{x/2}-e^{-x/2}}x\right]^2\tag3\\&=\left[\lim_{x\to0}\frac{e^x-e^{-x}}{2x}\right]^2\tag4\\&=\frac14\left[\lim_{x\to0}\frac{e^x-1}x-\lim_{x\to0}\frac{e^{-x}-1}x\right]^2\tag5\\&=\frac14\left[\lim_{x\to0}\frac{e^x-1}x-\lim_{x\to0}\frac{e^x-1}{-x}\right]^2\tag6\\&=\left[\lim_{x\to0}\frac{e^x-1}x\right]^2\tag7\\&=1^2=1\tag8\end{align}$$

Thus,

$$2L=1\implies L=\frac12$$

Steps:

$(1):$ Simply $(e^x-1-x)+(e^{-x}-1+x)=e^x+e^{-x}-2$

$(2):$ Turns out to be a perfect square. Who knew? ;)

$(3):$ We can take the square out.

$(4):$ Let $x\mapsto2x$.

$(5):$ Factor out the $1/2$ (which gets squared) and then split the limit into two, using: $e^x-e^{-x}=(e^x-1)-(e^{-x}-1)$.

$(6):$ Let $x\mapsto-x$ in the second limit.

$(7):$ Combine the resulting limits into one.

$(8):$ Use the well-known and probably given that $\lim_{x\to0}\frac{e^x-1}x=1$.

edited Aug 06 '17 at 23:58

answered Aug 06 '17 at 23:45

Simply Beautiful Art

74,685

You used $\lim_0\frac {e^x-1}{x}=1$. – hamam_Abdallah Aug 06 '17 at 23:49
What is $L$ in your first step? – Janitha357 Aug 06 '17 at 23:50
@Salahamam_Fatima Yes? – Simply Beautiful Art Aug 06 '17 at 23:51
@Janitha357 $L$ is the limit, given by$$L=\lim_{x\to0}\frac{e^x-1-x}{x^2}$$ – Simply Beautiful Art Aug 06 '17 at 23:52
@SimplyBeautifulArt How did you know that it existed? – Janitha357 Aug 06 '17 at 23:53
@SimplyBeautifulArt Step 8 is Taylor or l'Hospital. – hamam_Abdallah Aug 06 '17 at 23:54
@Janitha357 Presumptuously :-) A formal proof of the existence of the limit can be managed easily through series, which more or less amount to applying the MVT. – Simply Beautiful Art Aug 06 '17 at 23:54
@Salahamam_Fatima Both of those are circular ways to find $\lim_{x\to0}\frac{e^x-1}x$. This limit is usually given as the definition of $e$, or follows from the definition of $e$. – Simply Beautiful Art Aug 06 '17 at 23:55
SHOOT. I forgot I was only supposed to give out hints! Oh well... I mean, not quite sure how to formulate this into a hint anyways... – Simply Beautiful Art Aug 06 '17 at 23:58
@SimplyBeautifulArt Neat (+1). The last step could be also done by using $,e^x \ge 1+x,$ which has several proofs that are light on calculus. Then squeeze $,x \le e^x - 1 \le x /(1-x),$ around $,x = 0,$. – dxiv Aug 07 '17 at 00:18
@dxiv What definition of $e^x$ do you happen to be using? :-) – Simply Beautiful Art Aug 07 '17 at 00:21
@SimplyBeautifulArt For $,e^x \ge 1+x,$ it's enough to use $,e^x = \lim (1+x/n)^n,$ plus Bernoulli's. – dxiv Aug 07 '17 at 00:22
Fair enough (though I still think limits such as the one I used should be taken as givens for these sorts of questions) – Simply Beautiful Art Aug 07 '17 at 00:24
One approach is to use the definition $e^{x} =\lim_{n\to\infty} \left(1+\dfrac{x}{n}\right)^{n}$ and show that the limit in question for $x\to 0^{+}$ is $1/2$. The case $x\to 0^{-}$ can be deduced from the case of $x\to 0^{+}$ by using $e^{x} e^{-x} =1$. – Paramanand Singh Aug 07 '17 at 04:49
See this answer https://math.stackexchange.com/a/1565665/72031 – Paramanand Singh Aug 07 '17 at 04:56

score 2 · Answer 2 · answered Aug 06 '17 at 23:14

2

Hint Use the Taylor Series expansion of $e^{x}$, then expand and cancel like-terms.

answered Aug 06 '17 at 23:14

JohnColtraneisJC

1,919

Solving $\lim_{x\to0} \frac{e^x-1-x}{x^2}$

2 Answers2

No series or L'Hospital's required! "DISCLAIMER": This is not a hint.