1. ## two limit problems

please, Im having trouble finding these as x goes to infinity

$sqrt(2x+1)/x+4$ limit is 0 and $sqrt(x^2 + 2x) - x$ limit is 1

for the first I take both sides by 1/x but end up getting $sqrt(2)/1$ and for the second im unsure. i have the answers because of the back of the book but I dont know how to get them

2. $\frac{\sqrt{2x+1}}{x+4} =$

$\frac{\frac{\sqrt{2x+1}}{x}}{\frac{x+4}{x}} =$

$\frac{\frac{\sqrt{2x+1}}{\sqrt{x^2}}}{\frac{x+4}{x }} =$

$\frac{\sqrt{\frac{2}{x}+\frac{1}{x^2}}}{1 + \frac{4}{x}}$

now take the limit.

$(\sqrt{x^2 + 2x} - x) \cdot \frac{\sqrt{x^2 + 2x} + x}{\sqrt{x^2 + 2x} + x} =$

$\frac{(x^2 + 2x) - x^2}{\sqrt{x^2 + 2x} + x} =$

$\frac{2x}{\sqrt{x^2 + 2x} + x} =$

$\frac{2}{\sqrt{1 + \frac{2}{x}} + 1}$

now take the limit.

3. thank you that helped me a lot

4. Skeeter,

Could you use l'Hopital's Rule for the first problem instead? I tried it and got the same answer.
Thanks!

5. yes, L'Hopital will work on that problem.

however, consider this limit ...

$\lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}$

it's easy to see the limit is 1, but try using L'Hopital and see what happens.

6. The $x^2$ and the square root in the denominator is what causes problems using l"Hopital for this one right?