Question

[3-(5+x)^1/2] / (x-4). Find limit as X approaches 4

Answer 1

lim_(x->4) (3-sqrt(x+5))/(x-4)
Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->4) (3-sqrt(5+x))/(-4+x) = lim_(x->4) (( d(3-sqrt(5+x)))/( dx))/(( d(-4+x))/( dx)):
= lim_(x->4) -1/(2 sqrt(x+5))
Factor out constants:
= -1/2 (lim_(x->4) 1/sqrt(x+5))
The limit of a quotient is the quotient of the limits:
= -1/(2 (lim_(x->4) sqrt(x+5)))
Using the power law, write lim_(x->4) sqrt(x+5) as sqrt(lim_(x->4) (x+5)):
= -1/(2 sqrt(lim_(x->4) (x+5)))
The limit of x+5 as x approaches 4 is 9:
= -1/6

Answer 2

can you write it down please? i can't understand it this way. I'd be grateful

Answer 3

and isn't L'hopital's rule applicable only when X -> 0 ?