Finding limit from Squeeze theorem: $\lim\limits_{n\to\infty} \left(\frac{2n-5}{3n+1}\right)^n$

Question

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.

Answer 1

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

Answer 2

We have that

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

Answer 3

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.