How do you simplify #(sqrt(4+h)-2)/h# ?

1 Answer
Feb 25, 2017

Answer:

#(sqrt(4+h)-2)/h = 1/(sqrt(4+h)+2)#

with exclusion #h != 0#

Explanation:

Given:

#(sqrt(4+h)-2)/h#

We find:

#(sqrt(4+h)-2)/h = ((sqrt(4+h)-2)(sqrt(4+h)+2))/(h(sqrt(4+h)+2))#

#color(white)((sqrt(4+h)-2)/h) = ((4+h)-4)/(h(sqrt(4+h)+2))#

#color(white)((sqrt(4+h)-2)/h) = color(red)(cancel(color(black)(h)))/(color(red)(cancel(color(black)(h)))(sqrt(4+h)+2))#

#color(white)((sqrt(4+h)-2)/h) = 1/(sqrt(4+h)+2)#

#color(white)()#
Footnote

This is the sort of expression you find involved in a limit problem, like:

What is: #lim_(h->0) (sqrt(4+h)-2)/h# ?

The tricky thing here is that both the numerator and denominator become #0# when #h = 0#, but we can find the limit using our simplification...

#lim_(h->0) (sqrt(4+h)-2)/h = lim_(h->0) 1/(sqrt(4+h)+2)#

#color(white)(lim_(h->0) (sqrt(4+h)-2)/h) = 1/(sqrt(4+0)+2)#

#color(white)(lim_(h->0) (sqrt(4+h)-2)/h) = 1/(2+2)#

#color(white)(lim_(h->0) (sqrt(4+h)-2)/h) = 1/4#