Question

The value of the limit $\lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x}$ is:

• A. $\frac{5}{4}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Hint:

For $\lim_{x\to 0} \frac{e^{ax} - e^{-bx}}{\sin cx}$, the value of the limit simplifies directly to $\frac{a+b}{c}$. Here, $\frac{3 - (-2)}{4} = \frac{5}{4}$.

Answer 1

The Correct Option is A

Step 1: Check the form.
Put $x = 0$ first. The top becomes $e^0 - e^0 = 0$ and the bottom becomes $\sin 0 = 0$. So it is the $\frac{0}{0}$ form, which needs a method like L'Hopital's rule.

Step 2: Recall L'Hopital's rule.
For a $\frac{0}{0}$ limit, we may differentiate the top and the bottom separately and then take the limit again.

Step 3: Differentiate the top.
The derivative of $e^{3x}$ is $3e^{3x}$ and of $-e^{-2x}$ is $+2e^{-2x}$.
\[ \text{top}' = 3e^{3x} + 2e^{-2x} \]

Step 4: Differentiate the bottom.
The derivative of $\sin 4x$ is $4\cos 4x$.
\[ \text{bottom}' = 4\cos 4x \]

Step 5: Put $x = 0$ again.
Now substitute zero into the new fraction. The exponentials all become $1$ and $\cos 0 = 1$.
\[ \frac{3(1) + 2(1)}{4(1)} = \frac{5}{4} \]

Step 6: State the answer.
So the limit equals $\frac{5}{4}$.
\[ \boxed{\dfrac{5}{4}} \]