Confused about a limit

Question

Evaluate$$\lim\limits_{x\to\infty}\left(\dfrac{x}{\sqrt{x^2+2x+2}}\right)^{2x}$$

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: $$\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e \ , \ where \ n=\frac{\sqrt{x^2+2x+2}}{x-\sqrt{x^2+2x+2}}$$

Answer 1

HINT

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

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

$$...=e^{\lim\limits_{x\to\infty}\frac{x-x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}{x\cdot\sqrt{1+\frac{2}{x}+\frac{2}{x^2}}}\cdot2x}=e^{-2}$$

indeed

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

since

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

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

Answer 2

Hint:

$$\lim\limits_{x\to\infty}\left(\dfrac{x}{\sqrt{x^2+2x+2}}\right)^{2x}$$

$$=\lim_{x\to\infty}\left(\dfrac{x^2}{x^2+2x+2}\right)^x$$

$$=\lim_{x\to\infty}\left(\dfrac{x^2+2x+2}{x^2}\right)^{-x}$$

$$=\left(\lim_{x\to\infty}\left(1+\dfrac{2x+2}{x^2}\right)^{x^2/(2x+2)}\right)^{-\lim_{x\to\infty}\dfrac{2x+2}x}$$

The inner limit converges to $e$

Can you handle the limit in the exponent?