# Thread: Limit of a sequence

1. ## Limit of a sequence

Stuck on number 37. Do I do lhopitals rule taking the derivative on the top and bottom seperately?

3. ## Re: Limit of a sequence

Originally Posted by minneola24
Stuck on number 37. Do I do lhopitals rule taking the derivative on the top and bottom seperately?
$\displaystyle \frac{2n}{\sqrt{n^2+1}}=\frac{2}{\sqrt{1+n^{-2}}}$

4. ## Re: Limit of a sequence

Yes the solutions manual wrote that! But how? The derivative in the bottom is not that, I got something different.

5. ## Re: Limit of a sequence

Originally Posted by minneola24
Yes the solutions manual wrote that! But how? The derivative in the bottom is not that, I got something different.
Look, to do these problems one need to know how to us middle-school algebra.
Divide both numerator and denominator by n.

Do you know how to use basic algebra?

6. ## Re: Limit of a sequence

Originally Posted by minneola24
Yes the solutions manual wrote that! But how? The derivative in the bottom is not that, I got something different.
Plato did not use L'Hospital's Rule there. As he stated, he simply divided the top and the bottom by n. This gives an expression of which the limit can easily be evaluated.

7. ## Re: Limit of a sequence

Am I really this out of it.. I'm not seeing how you divide a square root by n.

I guess I wasn't paying too much attention in middle school!

8. ## Re: Limit of a sequence

\displaystyle \displaystyle \begin{align*} \frac{2n}{\sqrt{n^2 + 1}} &= \frac{\frac{2n}{n}}{\frac{\sqrt{n^2+1}}{n}} \\ &= \frac{2}{\frac{\sqrt{n^2+1}}{\sqrt{n^2}}} \\ &= \frac{2}{\sqrt{\frac{n^2+1}{n^2}}} \\ &= \frac{2}{\sqrt{1 + \frac{1}{n^2}}} \end{align*}