Find $\lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right)$ without using the L'Hopital's rule or Taylor's series

Question

This limit is proposed to be solved without using the L'Hopital's rule or Taylor series: $$ \lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right). $$ I know how to calculate this limit using the Taylor expansion: $$ x^2 \left( e^{1/x}-e^{1/(x+1)} \right)= x^2 \left( 1+\frac1{x}+\frac1{2!}\frac1{x^2}+\ldots - 1-\frac1{x+1}-\frac1{2!}\frac1{(x+1)^2}+\ldots \right) $$ $$ =x^2 \left( \frac1{x(x+1)}+\frac1{2!}\frac{2x+1}{x^2(x+1)^2}+\ldots \right)= \frac 1{1+\frac1{x}}+\frac1{2!}\frac{2x+1}{(x+1)^2}+\ldots, $$ thus, the limit is equal to $1$.

I'm allowed to use the fact that $\lim_{x\to0}\frac{e^x-1}{x}=1$, but I don't know how to apply it to this problem.

Both terms in parentheses tend to $1$. And then you can replace them with their logs (justify this). This simplifies the problem a lot. — Paramanand Singh, Oct 18 '22 at 07:52

zkutch · Accepted Answer · 2022-10-17T03:09:30.990

4

$$\lim_{x\to\infty}x^2 \left( e^{\frac1x}-e^{\frac{1}{x+1}} \right)=\lim_{x\to\infty}x^2 e^{\frac{1}{x+1}}\left( e^{\frac1x-\frac{1}{x+1}}-1 \right)=\\ =\lim_{x\to\infty}x^2 \left( e^{\frac{1}{x(x+1)}}-1 \right)=\lim_{x\to\infty}x^2\frac{1}{x(x+1)} = 1$$ On last step we use $\lim_\limits{y\to0}\frac{e^y-1}{y}=1$ considering it with composition $y=\frac{1}{x(x+1)}$, knowing $\lim_\limits{x\to\infty}\frac{1}{x(x+1)}=0$. Exact conditions for composition/substitution you can find on my answer from Bringing limits inside functions when limits go to infinity

edited Oct 17 '22 at 03:09

answered Oct 17 '22 at 02:51

zkutch

13,410

Where does the last step come from, writing $e^{\frac{1}{x(x+1)}} - 1 = \frac{1}{x(x+1)}$? – Fred T Oct 17 '22 at 02:58
Yes, only I prefer to write it as $e^{1 \over{x(x+1)}} - 1 \over{x(x+1)}$$\to 1$. – zkutch Oct 17 '22 at 03:00
right, but where does that come from? – Fred T Oct 17 '22 at 03:02
Looks like L'hopital to me. – A. Thomas Yerger Oct 17 '22 at 03:08
@Fred T. Added little explanation on end. – zkutch Oct 17 '22 at 03:10
@A. Thomas Yerger. Formally here is used only $\lim_{x\to0}\frac{e^x-1}{x}=1$, allowed by OP and substitution linked in answer. – zkutch Oct 17 '22 at 03:11
In the course from which this problem came from, $\lim_{x\to0}\frac{e^x-1}{x}=1$ follows from $\lim_{x\to0}(1+x)^{(1/x)}=e$, which, in turn, follows from the theorem that states that $\lim_{n\to\infty}(1+1/n)^n=e$, where $e$ is some number, $2<e<3$ – mathstudent1 Oct 17 '22 at 03:11

score -2 · Answer 2 · answered Oct 17 '22 at 03:19

-2

$$ L=\lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right). $$ Let $t=1/x$, $\frac{1}{1+x}=t(1+t)^{-1}=t(1-t+t^2+O(t^3))$, then $$L=\lim_{t \to 0} \frac{e^t-e^{t-t^2+O(t^3))}}{t^2}=\lim_{t\to 0} e^{t}\frac{1-e^{-t^2}}{t^2}=\lim_{t\to 0}e^t \frac{1-1+t^2+O(t^4)}{t^2}=1.$$

answered Oct 17 '22 at 03:19

Z Ahmed

43,235

2

From the question: "without using the L'Hopital's rule or Taylor series". – Gary Oct 17 '22 at 03:27

Find $\lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right)$ without using the L'Hopital's rule or Taylor's series

2 Answers2