How do you determine the limit of $lim_{x to +infty} x/sqrt(1+x^2)$ and $lim_{x to -infty} x/sqrt(1+x^2)$ ?

Question

it is right to think that for x that approach to $\pm infty$ , $sqrt(x^2+1)$ it's approximable to $sqrt(x^2)$ that is equal to $|x|$ so i can simplify my limit in $x/|x|$ ?

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Answer 1 · 2018-03-10T00:06:27

We have that

$lim_(x->+oo) x/(sqrt(1+x^2))=lim_(x->+oo) x/(absx*sqrt(1+1/x^2))=1/sqrt(1+0)=1$

(Because $x->+oo$ we have $absx=x$ )

Also

$lim_(x->-oo) x/(sqrt(1+x^2))=lim_(x->-oo) x/(absx*sqrt(1+1/x^2))=1/-sqrt(1+0)=-1$

(Because $x->-oo$ we have $absx=-x$ )

Answer 2 · 2018-03-10T00:07:27

Please see below.

Your suggestion will work for this limit. Here is another (similar) approach.

For all $x != 0$ ,

$sqrt(1+x^2) = sqrt(x^2)sqrt(1/x^2+1) = absx sqrt(1/x^2+1)$

On the right, we have $x > 0$ so

$x/sqrt(x^2+1) = x/(xsqrt(1/x^2+1)) =1/sqrt(1/x^2+1)$

and

$lim_(xrarroo)x/sqrt(x^2+1) = lim_(xrarroo)1/sqrt(1/x^2+1) = 1/sqrt(0+1) = 1$

On the left, we have $x < 0$ so

$x/sqrt(x^2+1) = x/(-xsqrt(1/x^2+1)) =-1/sqrt(1/x^2+1)$

and

$lim_(xrarr-oo)x/sqrt(x^2+1) = lim_(xrarr-oo)-1/sqrt(1/x^2+1) = -1/sqrt(0+1) = -1$