Question

solving limits at infinity(x->infinity), I have a problem x/sqrt(x^2+1). I tried L'Hospital's rule, and derivative, after simplification, ends up being another form of x/x or infinity/infinity. Is this true? or have I missed something?

Answer 1

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

Answer 2

if so, then lhopital it again

Answer 3

You can perform the derivative again.

Answer 4

ok remember the |x| is x if x>0
|x| is -x if x<0
since x->positive infinity then we have |x|=x

Answer 5

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

Answer 6

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

Answer 7

\[\frac{1}{\sqrt{1+0}}=\frac{1}{\sqrt{1}}=\frac{1}{1}=1\]

Answer 8

thats without lhospital

Answer 9

but you use that guy if you want to

Answer 10

thank you, that was what I was doing wrong. I should have tried to simplify it without L'hospital If you use L'Hospital, because of the sqrt and chain rule, you always end up with continued -1/2 exponents.