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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

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

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  • $\begingroup$ 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
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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.)

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  • $\begingroup$ 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

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