How do you evaluate the following limit? $$ \lim_{x\to\infty}\left(\frac{x^2}{2x+1}\right)^{\frac1x} $$
I tried to make it approach the shape of the euler limit
$$ \lim_{x\to\infty}\left(1+\frac{1}{f(x)}\right)^{f(x)} $$ I added and subtracted 1, so that we get "1 + a function", I inverted this function, making 1 / (1 / f (x)). then to the exponent I multiplied and divided by the function, in order to obtain a limit like that of euler, raised to another function but I realized that it was not the solution, because f (x) did not go to infinity when x went to infinity.
Then I gave up and asked here.
We can use exponential trick in this way
$$\left(\frac{x^2}{2x+1}\right)^{1/x}=e^{\frac{\log\left(\frac{x^2}{2x+1}\right)}x}\to e^0=1$$
indeed by standard limit $\frac{\log x}x \to 0$ we have
$$\frac{\log\left(\frac{x^2}{2x+1}\right)}x=\frac{\log\left(\frac{x^2}{2x+1}\right)}{\frac{x^2}{2x+1}}\frac{x}{2x+1}\to 0\cdot \frac 12 =0$$
See also the related
A simple proof of $\lim_{n\to \infty} \frac{\ln n}{n}=0$ for students of a high school