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


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 $$ \lim_{u \to 0}\left(\frac{\log (1+u)-u}{u^2}\right)=\lim_{u \to 0}\left(\frac{\frac1{1+u}-1}{2u}\right)=\lim_{u \to 0}\left(\frac{\frac{-1}{(1+u)^2}}{2}\right)=-\frac12. $$

  • $\begingroup$ Thank you for this however i used it $\endgroup$ – zeraoulia rafik Oct 18 '15 at 14:12
  • $\begingroup$ After the first Hopital you are left with $\frac{1-(1+u)}{2u(1+u)}=-\frac1{2(1+u)}$, so that the result can be obtained without the second invocation of el Hop. $\endgroup$ – LutzL Oct 18 '15 at 15:13
  • $\begingroup$ @LutzL You are right! Usually, I don't use L'Hospital's rule: I find the Taylor expansion really efficient. Thank you for your remark! $\endgroup$ – Olivier Oloa Oct 18 '15 at 16:12
  • $\begingroup$ Additionally, there is not much difference between using l'Hopital and the Taylor expansion. They are very closely related as immidiate consequences of the mean value theorem, the Taylor expansion is only a little more work in the coefficients, but more powerful in the application. $\endgroup$ – LutzL Oct 18 '15 at 16:15
  • $\begingroup$ @LutzL Right. That's why I've started my answer with "an approach is...", I had not seen the OP special intention:) $\endgroup$ – Olivier Oloa Oct 18 '15 at 16:20

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.)

  • $\begingroup$ The point here is not to argue this is the most efficient approach (see @Olivier Oloa's answer with L'Hopital, or the -- not allowed here -- Taylor expansion approach) -- mostly that very often, there is a good known inequality somewhere that will do the trick. And gives slightly more than just the limit as well , since you get quantifiable bounds in terms of simpler functions. (Also that there are other ways than L'Hopital.) $\endgroup$ – Clement C. Oct 18 '15 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.