Question

lim (x to infinity)(square root of ((x^2)+1)-x))

Answer 1

\[\lim_{x\rightarrow \infty} \sqrt{x^2+1}-x\]

Answer 2

gimmick is multiply top and bottom by conjugate

Answer 3

get
\[\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\]
\[\frac{1}{\sqrt{x^2+1}+x}\] take limit get zero

Answer 4

thx

Answer 5

@satellite73: Hey!! I have posted a question, I would appreciate that you have a look at it :)

Answer 6

yw

Answer 7

k, as long as it has nothing to do with laplace transforms from which i know from nothing

Answer 8

Agreed satellite^^

Answer 9

Lol, No! It's a programming problem. And by the way, Laplace transform is pretty easy :D

Answer 10

anwar i dont see your question but i am not good at programming i can look though

Answer 11

i wish i had learned to program.
i should spend some time reading about laplace transforms too, big caesura in my education. looks ok, lots of formulas

Answer 12

Here http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e3f17420b8ba3ff6fa943af

Answer 13

\[\lim_{x \rightarrow \infty }\frac{1}{\sqrt{x^2+1}+x}*\frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}\]
\[\lim_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{x^2}}}{\sqrt{1+\frac{1}{x^2}}+\frac{x}{\sqrt{x^2}}}\]
|x|=x when x>0
\[\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}+\frac{x}{x}}=\frac{0}{\sqrt{1+0}+1}=0\]