1

I'd like a general solution too, if possible!

Also, why is $$\lim_{n\to\infty}n\sin\left(\frac1n\right)=1?$$

Is it because $\sin \left(\frac1n\right)$ tends to $0$ at a decreasing rate compared to the regular rate at which $n$ tends to infinity?

Thanks.

Shaun
  • 44,997
Rascalniikov
  • 113
  • 3
    $$\frac{3^n}{2^n+3^n} = \frac{1}{(2/3)^n+1}$$ – peterwhy May 31 '14 at 14:33
  • 1
    @Shaun: your edit left out the n before the sine term. That's crucial. Tried editing it in, but it's less than 6 characters and the software won't let me. – Deepak May 31 '14 at 14:38
  • @Deepak Sorry. Of course, yeah. I'll correct it :) – Shaun May 31 '14 at 14:39
  • 2
    Keep questions distinct in future. Even if they both correspond to analysis, it is better to have two questions than just one. – Chinny84 May 31 '14 at 14:58

3 Answers3

2

Both of these limits can be handled with some rewriting -- for the first one: $$ \frac{3^n}{2^n+3^n}=\frac{1}{(2/3)^n+1} $$ Now look at what happens to $(2/3)^n$ as $n\to\infty$.

For $n\cdot\sin(1/n)$, again rewrite: $$ n\cdot\sin(1/n)=\frac{\sin(1/n)}{1/n} $$ and now look at what happens to $1/n$ as $n\to\infty$...

user128390
  • 809
1

hint: divide through $3^n$. What do you get?

Alex
  • 19,262
0

I believe you are mistaken, $\lim_{n\to\infty}\sin(1/n)=0$, and $\lim_{n\to\infty}\cos(1/n)=1$, this is because $\lim_{n\to\infty}1/n=0$, and $\sin(0)=0,\cos(0)=1$

Now lets approach the first question, simply divide through by $3^n$ to obtain:

$\lim_{n\to\infty}\frac{3^n}{2^n+3^n}=\lim_{n\to\infty}\frac{1}{(\frac{2}{3}^n+1)}=1$

Ellya
  • 11,783