AP Calculus AB Exam Question:

Find the limit algebraically:

Answer and Explanation:

To find the limit algebraically, we can start by multiplying and dividing the expression inside the limit by 3:

Next, we can simplify the expression inside the limit using the fact that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. Applying this result:

Applying the limits separately, we have:

We can evaluate each limit separately. First, for the left limit, we recognize that the expression inside the limit $\lim_{x \to 0} \left(\frac{3 \sin(3x)}{3x}\right)$ is of the form $\left(\frac{f(x)}{g(x)}\right)$, where $f(x) = 3 \sin(3x)$ and $g(x) = 3x$. Since $\lim_{x \to 0} \frac{f(x)}{g(x)} = 1$, we have:

For the right limit, we can express $\lim_{x \to 0} \left(\frac{1}{2x}\right)^{1/x}$ in the form $a^{1/x}$ by raising both the numerator and denominator to the $x$th power:

Now, we recognize that the expression inside the limit $\lim_{x \to 0} \left(\frac{1}{2x}\right)^x$ is of the form $a^x$, where $a = \frac{1}{2x}$. Using the property $\lim_{x \to 0} a^x = 1$, we can evaluate the limit as:

Finally, multiplying the left and right limits:

Therefore, the limit of the given expression as $x$ approaches 0 is 1.