If f′ is continuous, f(2)=0, and f′(2)=12, evaluate the limit:

Question

If f′ is continuous, f(2)=0, and f′(2)=12, evaluate the limit:

lim f(2+3x)+f(2+5x) / x

x--> 0

Answer 1

From your question it's hard to tell if you meant that "f(2+3x) + f(2+5x)" is all in the numerator or if "f(2+3x)" is not a part of the fraction. For the remainder of this answer, I'll assume both parts are in the numerator.

First observe that f(2+3x)=0 when x=0 because it is given to us in the problem that f(2) = 0. Similarly, f(2+5x) = 0 for the same reasons. Thus, we are evaluating a limit which has an indeterminate form. In other words, when we look for the limit, we can't find it directly because we have 0/0. So, we can use L'Hospital's Rule--we can take the derivative of the top and the derivative of the bottom.

Be careful! Taking the derivative of the top part is tricky because we have to employ the chain rule. Remember to take the derivative of the inside and multiply it by the derivative of the outside. Thus, the derivative of f(2+3x) will be f ' (2+3x) times 3. Similarly, the derivative of f(2+5x) will be f ' (2+5x) times 5.

Taking the derivative of the denominator is very straight forward. The derivative of x is just 1.

So our limit from the initial problem now becomes this:

"The limit as x approaches zero of (3f ' (2+3x) + 5f ' (2+5x))/1."

This limit can be evaluated directly by simply plugging in x=0. So we obtain that the limit as x approaches zero is equivalent to 3f ' (2) + 5f ' (2) or 8 f ' (2). Since it is given to us that f ' (2) = 12, the limit ultimately equals 8 * 12 which is 96.