Finding limit from Squeeze theorem.

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.

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

$$0<\left(\frac{2n-5}{3n+1}\right) ^{n}<\left(\frac{2}{3}\right)^n,$$ so $$\lim_{n\to\infty} \left(\frac{2n-5}{3n+1}\right) ^{n}=0.$$

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

We have that

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

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.

• Can you clarify what's wrong with my L'Hospital rule? – Sahil Silare Aug 26 '18 at 12:31
• @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. – J.G. Aug 26 '18 at 12:58