Why is the limit of this function tending to 1?

Question

$$ \lim_{x\to \infty} \left(\frac{1}{(x^2+x)\left(\ln\frac{x+1}{x}\right)^2}\right) $$

I know the answer is 1, but why does it tend to 1? Can you manipulate the function and the "$\ln$" to make it obvious?

Much appreciated.

Answer 1

Hint: $$\log (\frac{x+1}{x})=\log (1+\frac{1}{x})$$ and with $\frac{1}{x}\to 0$ we get $$\log (1+\frac{1}{x})\to\frac{1}{x}$$

Answer 2

$$ \lim_{x\to \infty} \left(\frac{1}{(x^2+x)\left(\ln\frac{x+1}{x}\right)^2}\right) $$

Use $u=\frac{1}{x}$

So we have $$\lim_{u\to 0} \left(\frac{u^2}{(1+u)\left(\ln\left[1+u\right]\right)^2}\right)$$ $$=\frac{1}{\lim_{u\to 0}(1+u)}\cdot \frac{1}{\lim_{u\to 0}\frac{\left(\ln\left[1+u\right]\right)^2}{u^2}} $$ $$=\frac{1}{\lim_{u\to 0}(1+u)}\cdot \frac{1}{\left(\lim_{u\to 0}\frac{\ln(1+u)}{u}\right)^2}$$ $$= \frac{1}{1+0}\cdot \frac{1}{1}$$ $$=1$$