How do you find the limit of #x / sqrt(x^2-x)# as x approaches infinity?
1 Answer
Jul 28, 2016
Explanation:
When
So:
#lim_(x->oo) x/sqrt(x^2-x)#
#=lim_(x->oo) x/sqrt(x^2(1-1/x))#
#=lim_(x->oo) color(red)(cancel(color(black)(x)))/(color(red)(cancel(color(black)(x))) sqrt(1-1/x))#
#=lim_(x->oo) 1/sqrt(1-1/x)#
#= 1/sqrt(1 + 0) = 1#