## anonymous 4 years ago find the limit please $\lim_{x \rightarrow infinity}\sqrt{x^2+3x+1}-x$

1. myininaya

you should put a 1 underneath that and rationalize the numerator by multiplying top and bottom by the conjugate of the top

2. myininaya

$\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+3x+1}-x}{1} \cdot \frac{\sqrt{x^2+3x+1}+x}{\sqrt{x^2+3x+1}+x}$

3. myininaya

$\lim_{x \rightarrow \infty}\frac{(x^2+3x+1)-x^2}{\sqrt{x^2+3x+1}+x}$

4. myininaya

$\lim_{x \rightarrow \infty}\frac{3x+1}{\sqrt{x^2+3x+1}+x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}$

5. myininaya

|x|=x if x>0 which it is since x gets positive large

6. myininaya

$\lim_{x \rightarrow \infty}\frac{3+\frac{1}{x}}{\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+\frac{x}{x}}$

7. myininaya

$\frac{3+0}{\sqrt{1+0+0}+1}$

8. anonymous

thanks once again

9. myininaya

lol np

10. anonymous

this is the last one, can you help me with it please

11. anonymous

Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity

12. anonymous

$\lim_{x \rightarrow +\infty} (x^2+x) / (x+1) = +\infty$$\lim_{x \rightarrow -\infty}(x^2 +x)/(x^2+1) =1$