AP Calculus AB Exam Question

Suppose $f(x) = \frac{\ln(x)}{x}$ and $g(x) = \frac{\cos(x)}{x^2}$. Use L'Hôpital's Rule to find the limit as $x$ approaches 0 of $\frac{f(x)}{g(x)}$.

(a) Find the derivatives of $f(x)$ and $g(x)$.

(b) Apply L'Hôpital's Rule to find $\lim_{x\to 0} \frac{f(x)}{g(x)}$.

(c) Determine whether the given limit is finite or infinite. If it is finite, find its value.

Answer

(a) First, let's find the derivatives of $f(x)$ and $g(x)$.

Derivative of $f(x)$: [ f'(x) = \frac{d}{dx}\left(\frac{\ln(x)}{x}\right) ] Applying the quotient rule: [ f'(x) = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} ] Simplifying: [ f'(x) = \frac{1 - \ln(x)}{x^2} ]

Derivative of $g(x)$: [ g'(x) = \frac{d}{dx}\left(\frac{\cos(x)}{x^2}\right) ] Again applying the quotient rule: [ g'(x) = \frac{-\sin(x) \cdot x^2 - \cos(x) \cdot 2x}{x^4} ] Simplifying: [ g'(x) = \frac{-x^2\sin(x) - 2x\cos(x)}{x^4} ] [ g'(x) = \frac{-x\sin(x) - 2\cos(x)}{x^3} ]

(b) We will apply L'Hôpital's Rule to find $\lim_{x\to 0} \frac{f(x)}{g(x)}$.

Using L'Hôpital's Rule: [ \lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f'(x)}{g'(x)} ] Substituting the derivatives: [ \lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{\frac{1 - \ln(x)}{x^2}}{\frac{-x\sin(x) - 2\cos(x)}{x^3}} ] [ = \lim_{x\to 0} \frac{x^3(1 - \ln(x))}{x^2(-x\sin(x) - 2\cos(x))} ] [ = \lim_{x\to 0} \frac{x(1 - \ln(x))}{-x\sin(x) - 2\cos(x)} ]

Now, substitute $x = 0$ into the expression, we get: [ \lim_{x\to 0} \frac{f(x)}{g(x)} = \frac{0(1 - \ln(0))}{-0\sin(0) - 2\cos(0)} ] [ = \frac{0(-\infty)}{-0 - 2(1)} = \frac{0}{-2} = 0 ]

(c) Since we found that the limit is finite, the value of the limit as $x$ approaches 0 of $\frac{f(x)}{g(x)}$ is 0.