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I recently came across a problem which stated to find the limit of an equation through the Squeeze theorem, $$\lim_{n\to\infty} \left(\frac{2n-5}{3n+1}\right) ^{n}$$ My approach: I did the question with L'Hospital's Rule just for the sake of finding the limit,

$$\log (L) = n(\log(2n-5) - \log(3n+1))$$ $$ \log(L) = \frac{\log(2n-5) - \log(3n+1)}{\frac{1}{n}}$$ By differentiating, $$ \log(L) = \frac{\frac{2}{2n-5}-\frac{3}{3n+1}}{\frac{-1}{n^2}}$$ $$ \log(L) = -\frac{17}{12}$$ $$ L = e^{-\frac{17}{12}}$$ This was the limit obtained by me. But I wasn't able to approach through Squeeze Theorem.

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    $\begingroup$ Are you sure of it? $\endgroup$ – Dr. Sonnhard Graubner Aug 26 '18 at 12:23
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    $\begingroup$ Is'nt your general term $<(2/3)^n$? $\endgroup$ – Lord Shark the Unknown Aug 26 '18 at 12:24
  • $\begingroup$ $\log(L) = \frac{\frac{2}{2n-5}-\frac{3}{3n+1}}{\frac{-1}{n^2}} \implies log(L) \to - \infty$ $\endgroup$ – ab123 Aug 26 '18 at 12:26
  • $\begingroup$ @ab123 Oh so we can't apply L'Hospital's rule twice? $\endgroup$ – Sahil Silare Aug 26 '18 at 12:28
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    $\begingroup$ The limit is $0$. $\endgroup$ – Anastassis Kapetanakis Aug 26 '18 at 12:28
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$$0<\left(\frac{2n-5}{3n+1}\right) ^{n}<\left(\frac{2}{3}\right)^n,$$ so $$\lim_{n\to\infty} \left(\frac{2n-5}{3n+1}\right) ^{n}=0.$$

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    $\begingroup$ In determining that the limit was "$-\frac{17}{12}$" you appear to have taken the limit as n goes to 0, not infinity. $\endgroup$ – user247327 Aug 26 '18 at 12:31
  • $\begingroup$ Sir can't the lower bound be $\left(\frac{-3}{4}\right)^{n}$ ? $\endgroup$ – Sahil Silare Aug 26 '18 at 13:44
  • $\begingroup$ no, when $n$ is even, $\left(\frac{-3}{4}\right)^{n}>\left(\frac{2}{3}\right)^n$. $\endgroup$ – Riemann Aug 26 '18 at 13:52
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We have that

$$0\le\left(\frac{2n-5}{3n+1}\right) ^{n}\le\left(\frac{2n-5+5}{3n+1-1}\right) ^{n}=\left(\frac{2}{3}\right) ^{n}\to 0$$

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I think you've tried to use L'Hopital's rule when only the denominator, not also the numerator, has $n\to\infty$ limit $0$. That's spurious. The correct analysis is $\ln L\to -\infty,\,L\to 0$ because of the asymptotic $(2/3)^n$ behaviour.

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  • $\begingroup$ Can you clarify what's wrong with my L'Hospital rule? $\endgroup$ – Sahil Silare Aug 26 '18 at 12:31
  • $\begingroup$ @SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating. $\endgroup$ – J.G. Aug 26 '18 at 12:58

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