Question

as the limit approaches infinity (sqrt(x^2 +4x +1))-x

Answer 1

multiply and divide by
\[\sqrt{x^2+4x-1}+x\]

Answer 2

So I did that. What next?

Answer 3

what did you get?

Answer 4

can you then use L'Hopital's?

Answer 5

Oh I typed part of that wrong the -1 under the root should be positive

Answer 6

What is that?

Answer 7

Did you get this??
\[\lim_{x \rightarrow \infty}\frac{ 2x^2+4x+1 }{ \sqrt{x^2+4x+1}-x }\]
If so divide denominator and numerator by x^2

Answer 8

I got 4x\[4x + 1 \div \sqrt{x^2 + 4x + 1} +x\]

Answer 9

The x's cancel out, not add together

Answer 10

The answer is 2..

Answer 11

Ya... you are right I am sorry...
Now divide denominator and numerator by x

Answer 12

\[\lim_{x \rightarrow \infty}\left( \sqrt{x^2+4x+1}-x \right)=\lim_{x \rightarrow \infty}\left( \sqrt{x^2+4x+1}-x \right)\frac{\left( \sqrt{x^2+4x+1}+x \right)}{\left( \sqrt{x^2+4x+1}+x \right)}\]
\[=\lim_{x \rightarrow \infty}\frac{\left( \sqrt{x^2+4x+1} \right)^2-x^2}{\left( \sqrt{x^2+4x+1} +x\right)}=\lim_{x \rightarrow \infty}\frac{x^2+4x+1-x^2}{\left( \sqrt{x^2+4x+1} +x\right)}=\lim_{x \rightarrow \infty}\frac{4x+1}{\left( \sqrt{x^2+4x+1} +x\right)}\]
\[=\lim_{x \rightarrow \infty}\frac{\frac{4x}{x}+\frac{1}{x}}{\left( \sqrt{1+\frac{4}{x}+\frac{1}{x^2}} +1\right)}=\lim_{x \rightarrow \infty}\frac{4+\frac{1}{x}}{\left( \sqrt{1+\frac{4}{x}+\frac{1}{x^2}} +1\right)}=\frac{4}{1+1}=2\]

Answer 13

Thanks!