Calculus problem?

First off, try apply the limit value. You get 0/0 which is undefined.

So try using L'hopital's rule by taking the derivative of the top and bottom and reapplying the limit

The derivative of the top is 1

The derivative of the bottom is (1/2)(84 + x)^(-1/2)

Reapply the limit

The top stays constant at 1

The bottom becomes (1/2)(81)^(-1/2) = 2/9

1/ (2/9) = 9/2

When you plug in negative three, you end up with the fraction 0/0 (which is an indeterminate). Therefore, you need to use L'Hospital's rule, which states that you must take the derivative of the numerator and the derivative of the denominator when the limit is indeterminate.

Taking the derivatives of the numerator and denominator separately (do not use quotient rule), you get:

lim (x -> -3) of 1 / (0.5*(84+x)^(-0.5))

Plugging in negative three, you obtain an answer of 18.

Where both the top and bottom of the fraction tend to 0 at the limit you can use L'Hopital's rule and do the limit of the differential of the top with respect to x over the differential of the bottom to get the result

d(x+3)/dx=1

d[(84+x)^0.5-9]/dx=0.5*(84+x)^-(0.5)

=1/18

the limit is therefore 1/(1/18)=18

Find derivative of top and bottom then plug in x