-2

I have to find the following limits:

a) $\displaystyle\lim_{x\to \infty}x\left(\left(1+\frac {1}{x}\right)^x-e\right)$.

b)$\displaystyle\lim_{x\to 0^+}\left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$

I had thought to solve this using l'Hôpital's Rule.

But for a I obtained that of recurring type. So I could not proceed through that.

As I had put $t=1/x$, so $\frac {(1+t)^{1/t}-e}{t}$ as $t \to 0$

By applying the Rule I got $\frac {-(1+t)^{1/t}ln(1+t)}{t^2}$. Again applying the Rule I obtained a recurring cycle.

Is there another way to do this or did I do any mistake?

b) For b I obtained an answer using the series of $\sin x$ and using the formula of the exponential. That leads to my answer as $e^{-1/6}$. Is it possible to do this example using L'Hôpital's Rule efficiently?

Community
  • 1
Curious student
  • 6,583
  • use the facts that $f(x)^{g(x)} = e^{g(x) \ln f(x)}$ and $\lim e^{f(x)} = e^{\lim f(x)}$ – D F May 06 '18 at 14:50
  • Deleting shortly after getting an answer raises eyebrows here. And denies the answerers a shot at a reward in terms of upvotes. Not sure that this is a great question, but that other consideration trumps that. – Jyrki Lahtonen May 06 '18 at 17:37
  • 1
    All My Respected Teachers on MSE, I am a ordinary IQ student ,But I wanted Whatever math came across [This is not exaggeration but my dream].Although I am member of this site Form last one month ,But I got Utility of this site just 10 days ago .Form that I Started using this site ,Whenever I had stuck in some problem I am posting Here .Actually I like this process of Learning ... Really I had to thank all Community for that ..... – Curious student May 07 '18 at 04:24
  • 1
    @ Jyrki Lahtonen♦ ,Respected Sir, My Intentesion was not that .I had posted that question before That I tried L'Hospital Rule But I wanted to find some Efficient method (if it exist )Because by that method I have to Do recurrisive work....
    Extremly Sorry For that .As Some downvote arrives for question I thought This question is not worth asking ... As I already mentioned I am A ordinary IQ Student I have not even ability to judge the question that it is of worth asking or not .That's why I had deleted .....Please Understand Me .....And Answer Posted By one of My teacher same as I had     – Curious student May 07 '18 at 04:30
  • 1
    I had already mentioned in post – Curious student May 07 '18 at 04:30
  • If you don't have the ability to judge, leave it to the community and the system. They will handle it. If you question gets closed without upvotes or an upvoted/accepted answer , it will be automatically deleted as RemoveAbandonedClosed by a script. – GNUSupporter 8964民主女神 地下教會 May 08 '18 at 12:36
  • a) appears here https://math.stackexchange.com/questions/73243/limit-of-x-left-left1-frac1x-rightx-e-right-when-x-to-infty – Arnaud D. Nov 28 '19 at 12:18
  • and b) appears here https://math.stackexchange.com/questions/700100/evaluating-lim-x-rightarrow-0-left-frac-sinxx-right-frac1x2 – Arnaud D. Nov 28 '19 at 12:36

1 Answers1

1

The idea of doing $x=1/t$ is good: $$ \lim_{t\to0^+}\frac{(1+t)^{1/t}-e}{t} $$ If $f(t)=(1+t)^{1/t}$, then, for $t>0$, $$ f'(t)=(1+t)^{1/t}\frac{t-(1+t)\log(1+t)}{t^2(1+t)} $$ Since $\displaystyle\lim_{t\to0^+}\frac{(1+t)^{1/t}}{1+t}=e$, you are reduced to computing $$ \lim_{t\to0^+}\frac{t-(1+t)\log(1+t)}{t^2}= \lim_{t\to0^+}\frac{1-1-\log(1+t)}{2t}=-\frac{1}{2} $$

For the second limit, compute instead the limit of the logarithm, that is, $$ \lim_{x\to0^+}\frac{\log\dfrac{\sin x}{x}}{x^2} $$ Since $\frac{\sin x}{x}=1-\frac{x^2}{6}+o(x^2)$, you get $$ \lim_{x\to0^+}\frac{-x^2/6+o(x^2)}{x^2}=-\frac{1}{6} $$ So your given limit is $e^{-1/6}$.

You can also do it with l'Hôpital, but it's not as pretty: $$ \lim_{x\to0^+}\frac{\dfrac{\cos x}{\sin x}-\dfrac{1}{x}}{2x}= \lim_{x\to0^+}\frac{x\cos x-\sin x}{2x^2\sin x}= \lim_{x\to0^+}\frac{-x\sin x}{4x\sin x+2x^2\cos x}= -\frac{1}{2}\lim_{x\to0^+}\frac{\sin x}{2\sin x+x\cos x}= -\frac{1}{2}\lim_{x\to0^+}\frac{\cos x}{3\cos x-x\sin x} $$

egreg
  • 238,574