How to apply l'Hopital to calculate the limit of the function (x^2-x-6)/(x+2) x-->-2
To calculate the limit, we'll substitute x by -2 in the expression of the function.
lim f(x) = lim ( 4 + 2 -6 )/( -2 + 2 )
lim ( x^2 - x -6 )/( x+2 ) = (6-6)/(-2+2) = 0/0
We notice that we've obtained an indetermination case, 0/0 type.
We'll apply l'Hospital's rule, by differentiating separately the numerator and denominator.
We'll differentiate the numerator:
( x^2 - x -6)' = 2x - 1
We'll differentiate the denominator:
(x+2)' = 1
We'll re-write the limit:
lim ( x^2 - x -6 )/( x+ 2 ) = lim (2x-1)
We'll substitute x by -2:
lim (2x-1) = 2*(-2) - 1 = -4-1
The limit of the function, when x->-2, is: lim (x^2-x-6)/(x+2) = -5.
In a limit `lim_(x->a) (f(x))/(g(x))` , if `(f(a))/(g(a)) = 0/0` or `(+-oo)/(+-oo)` , it is possible to apply l'Hospoital's rule and substitute `(f(x))/(g(x))` with `((f(x))')/((g(x))')`
In the given problem, the limit `lim_(x->-2)(x^2-x-6)/(x+2)` is required.
If we substitute x = -2, the result is `(4 + 2 - 6)/(-2 + 2) = 0/0` . This is an indeterminate form and we can apply l'Hospital's rule. Replace the numerator and denominator by their derivatives.
`lim_(x->-2) (2x - 1)/(1)`
Substituting x = -2 gives (2*-2 -1)/1 = -5
The required limit is `lim_(x->-2)(x^2-x-6)/(x+2) = -5`