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Answer 1 · 2016-10-20T22:45:23

For #x != 0#, we have

#x/sqrt(x^2+1) = x/sqrt(x^2(1+1/x^2))#

# = x/(sqrt(x^2)sqrt(1+1/x^2))#

Since #sqrt(x^2) = absx#, we have

#lim_(xrarroo) x/sqrt(x^2+1) =lim_(xrarroo)x/(xsqrt(1+1/x^2))#

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

# = 1/sqrt(1+0) = 1#

#y=1# is a horizontal asymptote on the right.

On the left, we have #x < 0#, so #sqrt(x^2) = -x# and

#lim_(xrarr-oo) x/sqrt(x^2+1) =lim_(xrarroo)x/(-xsqrt(1+1/x^2))#

# = lim_(xrarroo)-1/sqrt(1+1/x^2)#

# = -1/sqrt(1+0) = -1#

#y= -1# is a horizontal asymptote on the left.