How do you find the limit #(3+x^(-1/2)+x^-1)/(2+4x^(-1/2))# as #x->0^+#?

Question

1124 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T20:51:24

#oo#

#lim_(x to 0^+) (3+x^(-1/2)+x^-1)/(2+4x^(-1/2))#

#= lim_(x to 0^+) (3x+x^(1/2)+1)/(2x+4x^(1/2))#

#= lim_(x to 0^+) (sqrt x (3sqrt x+1)+1)/(sqrt(x) (2sqrtx+4))#

#= lim_(x to 0^+) color(red)(( 3sqrt x+1)/( 2sqrtx+4) )+color(blue)( (1)/(sqrt(x) (2sqrtx+4)))#

For #x "<<"1#, the red term is: #1/4#.

However, for the blue term:

#= lim_(x to 0^+) (1)/(sqrt(x) (2sqrtx+4)) = lim_(x to 0^+) (1)/(sqrt(x) * (4)) = oo#