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#