How do you simplify #(sqrt(4+h)-2)/h# ?
1 Answer
with exclusion
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)#
Footnote
This is the sort of expression you find involved in a limit problem, like:
What is:
The tricky thing here is that both the numerator and denominator become
#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#