How do you find #lim (2-sqrt(x+2))/(4-x^2)# as #x->2# using l'Hospital's Rule?

Question

3652 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T17:34:41

#lim_(x->2) frac( 2-sqrt(x+2)) (4-x^2) =1/16#

First we observe that:

#lim_(x->2) 2-sqrt(x+2) = 0#

#lim_(x->2) 4-x^2= 0#

So the limit:

#lim_(x->2) frac( 2-sqrt(x+2)) (4-x^2)#

is in the indeterminate form #0/0# and we can use l'Hospital's rule:

#lim_(x->2) frac( 2-sqrt(x+2)) (4-x^2) = lim_(x->2) frac (d/(dx) (2-sqrt(x+2))) (d/(dx)(4-x^2))#

Calculate the derivatives:

#d/(dx) (2-sqrt(x+2)) = -1/(2sqrt(x+2))#

#d/(dx)(4-x^2) = -2x#

So:

#lim_(x->2) frac( 2-sqrt(x+2)) (4-x^2) = lim_(x->2) 1/(4x(sqrt(x+2)))=1/16#