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

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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]}