How should I calculate $\lim_{n\to\infty}\left(\frac{2n-3}{3n+4}\right)^{n+1}$?

Question

I have been studying for my Analysis exam but I have problem that can't solve.

Original: $$\lim_{n\to\infty}\left(\dfrac{2n-3}{3n+4}\right)^{n+1}$$

What I did trying to solve it making it looks like a sequence to e: $$\lim_{n\to\infty}\left(\frac{3n+4}{3n+4}+\frac{-n-7}{3n+4}\right)^{n+1}$$ That made into: $$\lim_{n\to\infty}\left(1+\frac{-n-7}{3n+4}\right)^{n+1}$$

I don't know what to do with -n to dissappear from the equation.

And this is what I should get in the end

$$\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^{n} = e^a$$

I hope it is detailed enough.

Answer 1

[This question has been changed by OP. Orginally, it was about calculating $\lim_{n\to \infty} \biggr(\dfrac{-n-3}{3n+4}\biggr)^{n+1}$.]

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Your attempt is in the wrong track. The strategy should be very different from considering $$ \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\tag{1} $$ because they are two different types of problems. In your case $$ \frac{2n-3}{3n+4}\to\frac{2}{3}<1\quad \textrm{as } n\to\infty, $$ but in (1), $$ 1+\frac{1}{n}\to 1 \quad \textrm{as } n\to\infty $$ In the later case, you have the form $1^\infty$ while in your question, you don't.

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Note that $$ \left(\frac{2n-3}{3n+4}\right)^{n+1}=\exp\left((n+1)\cdot\log\biggr(\frac{2n-3}{3n+4}\biggr)\right). $$ What can you say about $$ \lim_{n\to\infty}(n+1)\log\biggr(\frac{2n-3}{3n+4}\biggr)? $$

Note in particular that $$ \lim_{n\to\infty}\log\biggr(\frac{2n-3}{3n+4}\biggr)=\log\biggr(\frac23\biggr)<0. $$

Answer 2

Observe:

$$\frac{n+3}{3n+4}=\frac13\,\frac{n+3}{n+\frac43}=\frac13\left(1+\frac{\frac53}{n+\frac43}\right)=\frac13+\frac5{9n+12}$$

and thus

$$1-\frac{n+3}{3n+4}=\frac23-\frac5{9n+12}=\frac23\left(1-\frac5{6n+8}\right)$$

Can you take it from here?

Answer 3

You have

$$\lim_{n\to \infty} \frac{-n-3}{3n+4}=-\frac 13.$$

So

$$\lim_{n\to \infty}\left( 1+\frac{-n-3}{3n+4}\right)=\frac 23.$$

Now you know because

$$0<\frac 23<1$$

that

$$\lim_{n\to\infty }\left(\frac 23\right)^n=0.$$

So the limit you are looking for is $0$.

Answer 4

I get the impression that you are not convinced of the fact that your limit is evidently 0, perhaps this alternative (longer) way can show you why. $$ \lim_{n\to\infty}\left(\dfrac{2n-3}{3n+4}\right)^{n+1}= $$ consider: $$ \frac{2n-3}{3n+4}=\frac{2n-3-\alpha(3n+4)+\alpha(3n+4)}{3n+4}=\frac{2n-3-\alpha(3n+4)}{3n+4}+\alpha $$ if $\alpha$ is chosen to be $\alpha\to\frac{2}{3}$ then by some elementary algebra: $$ \frac{2n-3}{3n+4}=\frac{2}{3}-\frac{17}{9n+12}=\frac{2}{3}\left(1-\frac{17}{6n+8}\right) $$ which means that your limit now is: $$\lim_{n\to\infty}\left(\frac{2}{3}\right)^{n+1}\left(1-\frac{17}{6n+8}\right)^{n+1}$$ now by a simple substitution $n\to \frac{1}{6} (17 t-8)$ you get: $$\lim_{n\to\infty}\left(\frac{2}{3}\right)^{n+1}*\lim_{t\to\infty}\left(1-\frac{1}{t}\right)^{\frac{1}{6} (17 t-2)}=e^{-17/6}\lim_{n\to\infty}\left(\frac{2}{3}\right)^{n+1}=0$$