How do I evaluate this without using taylor expansion :$\lim_{x \to \infty}x^2\log(\frac {x+1}{x})-x\ $?

Question

How do I evaluate this without using Taylor expansion?

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

Note: I used Taylor expansion at $z=0$ and I have got $\frac{-1}{2}$

Thank you for any help

Answer 1

An approach is to set $u:=\dfrac1x$, then you are looking for $$ \lim_{u \to 0}\left(\frac{\log (1+u)-u}{u^2}\right) $$ then you may conclude with L'Hospital's rule $$ \lim_{u \to 0}\left(\frac{\log (1+u)-u}{u^2}\right)=\lim_{u \to 0}\left(\frac{\frac1{1+u}-1}{2u}\right)=\lim_{u \to 0}\left(\frac{\frac{-1}{(1+u)^2}}{2}\right)=-\frac12. $$

Answer 2

Another option is to use the following (very convenient) inequality: $$ \frac{2x}{2+x} \leq \ln(1+x) \leq \frac{x}{\sqrt{1+x}} $$ which holds for $x \geq 0$. In your case, you have $\ln(1+\frac{1}{x})$, leading to $$ x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x \leq x^2\ln\!\left(1+\frac{1}{x}\right) -x \leq x^2\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x}}} - x. $$

• For the LHS: $$ x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x = x\left(\frac{2}{2+\frac{1}{x}} - 1\right) = x\left(\frac{-\frac{1}{x}}{2+\frac{1}{x}}\right) = \frac{-1}{2+\frac{1}{x}} \xrightarrow[x\to\infty]{} -\frac{1}{2}. $$

• And the RHS: $$\begin{align} x^2\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x}}} - x &= x\left(\frac{1}{\sqrt{1+\frac{1}{x}}} - 1\right) = x\left(\frac{1-\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}}}\right)\\ &= x\left(\frac{-\frac{1}{x}}{\sqrt{1+\frac{1}{x}}\left(1+\sqrt{1+\frac{1}{x}}\right)}\right)\\ &= \frac{-1}{\sqrt{1+\frac{1}{x}}\left(1+\sqrt{1+\frac{1}{x}}\right)} \xrightarrow[x\to\infty]{} -\frac{1}{2}. \end{align}$$ It only remains to invoke the squeeze theorem.

(I also strongly suggest bookmarking the "useful inequalities sheet" linked at the beginning.)