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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.

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    What was the last step before you got stuck? – 2017-01-08
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    I don't know what to do with -n. All other steps are clear. This picture made after I made some changes. – 2017-01-08
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    BTW, in that link that you have provided (and please refrain from doing so next time, just add it directly to your question) - there's absolutely no sequence, and no problem to solve. Just an expression. – 2017-01-08
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    What do you mean "all other steps"? There are no steps listed there either! – 2017-01-08
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    It's a sequence converging to e and I have to calculate limn->infinite and that expression. – 2017-01-08
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    Welcome to MSE! To learn how to type pretty math, please see [the MathJax Guide](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference), which allows you to write interesting things. $$\overbrace{\left(\ddot{\stackrel{\quad>}{\smile}}\right)}_{\begin{align}\hline\qquad\end{align}}$$Cheers! – 2017-01-08
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    It doesn't tend to $e$ – 2017-01-08
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    There's no sequence there, only an expression. If you meant to ask about $\lim\limits_{n\to\infty}$ of that expression, then please be advised that it is not shown in that linked image. And again - none of the steps that you have made in order to solve it are listed there either. – 2017-01-08
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    This sequence goes to zero. – 2017-01-08
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    This question has been changed by OP. Orginally, it was about calculating $\lim_{n\to \infty} \biggr(\dfrac{-n-3}{3n+4}\biggr)^{n+1}$. – 2017-01-08

4 Answers 4

0

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: $$\lim_{n\to\infty}\left(\frac{2}{3}\right)^{n+1}*\lim_{t\to\infty}\left(1-\frac{1}{t}\right)^{\frac{1}{6} (17 t-2)}=e^{-17/6}\lim_{n\to\infty}\left(\frac{2}{3}\right)^{n+1}=0$$

1

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?

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    It's also not the same, has no exponent. – 2017-01-08
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    @Rinner This is a *hint* and because of that the "can you take it from here?" Complete it and you'll see the limit is zero... – 2017-01-08
1

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 $0$.

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    That is not the case: It's not that simple: it is exponential. – 2017-01-08
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    Add parenthesis in second limit. – 2017-01-08
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    @Rinner You should really think a little more deeper before you write a comment. This answer, just as mine, requires from you thing, justify and complete. – 2017-01-08
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    @Rinner I agree with DonAntonio. – 2017-01-08
1

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

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.

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