# Thread: limit of a sequence

1. ## limit of a sequence

Here's another limit:

$\displaystyle \lim_{n\to \infty}\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}$

That's all I could get from it:

$\displaystyle \lim_{n\to \infty}\frac{\left(1+\frac{2}{n^2}\right)^{n^2}}{2 ^{n^2}\left(1+\frac{1}{2n^2}\right)^{n^2}}$

And everything would be quite fine if there was no $\displaystyle 2^{n^2}$.... Then the limit would be $\displaystyle e^\frac{3}{2}$, just like the answers say.... Am I wrong somewhere, or is there a mistake in this example?

Thank you in advance.

2. That expresion is equivalent to:
$\displaystyle \left[ {\frac{{\left( {1 + \frac{2}{{n^2 }}} \right)^{n^2 } }}{{\left( {1 + \frac{1}{{2n^2 }}} \right)^{n^2 } }}} \right]\left[ {\frac{1}{{2^{n^2 } }}} \right]$

What is the limit of a product?

3. It's equal to the product of the limits.... But the limit of $\displaystyle \frac{1}{2^{n^2}}$ when $\displaystyle n$ tends to infinity is zero, I believe; so the final result will be zero too?

4. Originally Posted by disclaimer
But the limit of $\displaystyle \frac{1}{2^{n^2}}$ when $\displaystyle n$ tends to infinity is zero, I believe; so the final result will be zero too?
Don't just believe it. Know that it is true.

5. Thank you. Now I guess that there's a mistake in my textbook, that's why I was a bit unsure. Thanks for clearing it up.

6. The bottom is always more than the top and it grows faster, so the limit goes to zero.