AP Calculus AB Exam Question

Let f(x) and g(x) be continuous functions on an interval containing x = a. Suppose that f(a) = g(a) = 0 and that g'(a) ≠ 0. Using L'Hôpital's Rule, determine the value of the limit:

lim(x→a) [f(x)/g(x)]

Answer with Step-by-Step Detailed Explanation

L'Hôpital's Rule states that if a limit of the form lim(x→a) [f(x)/g(x)] has the indeterminate form 0/0 or ±∞/±∞, then we can take the derivative of the numerator and denominator until we no longer have the indeterminate form.

Let's apply L'Hôpital's Rule to the given limit. We have:

lim(x→a) [f(x)/g(x)]

Since f(a) = g(a) = 0, we can rewrite the limit as:

lim(x→a) [f(x) - 0] / [g(x) - 0]

This gives us the indeterminate form 0/0. By applying L'Hôpital's Rule, we can take the derivative of the numerator and denominator. Differentiating both numerator and denominator, we get:

lim(x→a) [f'(x)/g'(x)]

Now, the limit is in a simplified form, and evaluating this limit will give us the desired value.

Taking the limit as x approaches a, we have:

lim(x→a) [f'(x)/g'(x)]

Substituting a into the expression, we get:

lim(x→a) [f'(a)/g'(a)]

Since g'(a) ≠ 0, we can evaluate this limit directly. Therefore, the value of the given limit is:

f'(a) / g'(a)

Hence, the value of the limit lim(x→a) [f(x)/g(x)] using L'Hôpital's Rule is f'(a) / g'(a).

Note: It is important to consider the given conditions, such as f(a) = g(a) = 0 and g'(a) ≠ 0, when applying L'Hôpital's Rule.