What is the limit of $sqrt[ x^2-1]/ sqrt[x^2+1]$ as x goes to infinity?

Question

7554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-15T10:15:16

The limit is $=1$

Let's rewrite the function

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

$=(cancelxsqrt(1-1/x^2))/(cancelxsqrt(1+1/x^2))$

$lim_(x->oo)1/x^2=0$

Therefore,

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

$=(sqrt(1-0))/(sqrt(1-0))$

$=1$

graph{(y-sqrt(x^2-1)/sqrt(x^2+1))(y-1)=0 [-1.15, 11.337, -1.79, 4.453]}

What is the limit of sqrt[ x^2-1]/ sqrt[x^2+1]sqrt[ x^2-1]/ sqrt[x^2+1] as x goes to infinity?