EDIT: Thank you for your answers and suggestions, I am sorry I didn't use MathJax from the beginning.

But if I may ask, why this doesn't led to the solution?: $$\lim\limits_{x\to\infty}\left(\frac{x}{\sqrt{x^2+2x+2}}\right)^{2x}=\lim\limits_{x\to\infty}\left(1+\frac{x-\sqrt{x^2+2x+2}}{\sqrt{x^2+2x+2}}\right)^{2x}=\lim\limits_{x\to\infty}\left(1+\frac{x-\sqrt{x^2+2x+2}}{\sqrt{x^2+2x+2}}\right)^{\frac{\sqrt{x^2+2x+2}}{x-\sqrt{x^2+2x+2}}\cdot\frac{x-\sqrt{x^2+2x+2}}{\sqrt{x^2+2x+2}}\cdot2x}=e^{\lim\limits_{x\to\infty}\frac{x-\sqrt{x^2+2x+2}}{\sqrt{x^2+2x+2}}\cdot2x}=e^{\lim\limits_{x\to\infty}\frac{x-x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}{x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}\cdot2x}=e^{0/\infty}=e^0=1$$ I used Euler's number: $$\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e \ , \ where \ n=\frac{\sqrt{x^2+2x+2}}{x-\sqrt{x^2+2x+2}}$$

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    $\begingroup$ write your formula here please! $\endgroup$ – Dr. Sonnhard Graubner Mar 5 '18 at 15:54
  • $\begingroup$ ... and use mathjax math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Lee Mosher Mar 5 '18 at 15:57
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    $\begingroup$ Welcome to stackexchange. Your question is attracting downvotes and votes to close instead of answers because you've made it much too hard for anyone here to help you. If you edit the question to include the actual text (not a picture) and explain where you think your error might possibly be in "the second way" maybe someone will take a look. Use mathjax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Ethan Bolker Mar 5 '18 at 15:58
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    $\begingroup$ You were going in the right direction covering the hard part of the question and then for the simpler part you made the wrong move. The limit of the exponent $(x-\sqrt{\dots}) \cdot 2x/\sqrt {\dots} $ is evaluated by multiplying numerator and denominator with conjugate $x+\sqrt{\dots} $. After thus you can easily see the exponent tends to $-2$ and the answer is $1/e^2$. $\endgroup$ – Paramanand Singh Mar 5 '18 at 16:58
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    $\begingroup$ You are not supposed to get frustrated while doing math. You are supposed to enjoy it (that's the secret to math which no one believes btw):) :) $\endgroup$ – Paramanand Singh Mar 5 '18 at 17:29


$$\left(\dfrac{x}{\sqrt{x^2+2x+2}}\right)^{2x}=e^{2x\log\left(\dfrac{x}{\sqrt{x^2+2x+2}}\right)}=e^{-2x\log\left( \sqrt{1+\frac2x+\frac2{x^2}} \right)}=e^{-x \log\left( 1+\frac2x+\frac2{x^2} \right)}$$

then use infinitesimal expansion for $y\to 0 \implies \log (1+y) \sim y$ for

$$\log\left( 1+\frac2x+\frac2{x^2} \right)\sim \frac2x+\frac2{x^2}$$

You can fix your work as follow



$$\frac{x-x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}{x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}\cdot2x=\frac{x-x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}{\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}\cdot2 \to -2$$


$$x-x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}\sim x-x-1-\frac1x\to-1$$

in the last we have used binomial expansion for $y\to 0 \implies (1+y)^n=1+ny$.

  • $\begingroup$ Thank you! A final question: what did I do wrong in my example? $\endgroup$ – MathNoob Mar 5 '18 at 17:01
  • $\begingroup$ @MathNoob It seems that in the final step you have an indeterminate form $0\cdot \infty$ $\endgroup$ – gimusi Mar 5 '18 at 17:07
  • $\begingroup$ @MathNoob I've added the way you can conclude by your work! $\endgroup$ – gimusi Mar 5 '18 at 17:13
  • $\begingroup$ Thank you so much for your effort you put in order to make me trying to understand. I am sorry if I am too dumb but can you explain a little more in depth why is $$x-x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}\ \ \ \sim\ x-x-1-\frac{1}{x}$$ my stupid brain thinks it should be $$x-x\cdot\sqrt{1}=x-x=0$$ and I kinda don't really get it with the binomaial expansion. Thank you! $\endgroup$ – MathNoob Mar 5 '18 at 17:29
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    $\begingroup$ Oooh, I get it now! I was going to the limit, while still $$using \ x \ instead \ of \infty$$ , which would give an indeterminate form $$\infty \ -\infty$$ Now I truly understand it! Thank you so much for your help! $\endgroup$ – MathNoob Mar 5 '18 at 17:41






The inner limit converges to $e$

Can you handle the limit in the exponent?

  • $\begingroup$ Thank you! I solved the problem in this way, the final result being $$\frac{1}{e^2}.$$ But I was wondering why in my example (without raising the fraction to the power of 2) , didn't get me the same answer. $\endgroup$ – MathNoob Mar 5 '18 at 17:02

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