The Limit — Question 8

Question 8

Evaluate the following limit: $$\lim_{x \to 0} \frac{e^{2x} - 1}{\sin(3x)}$$

Question 8 - Solution

We are asked to compute: $$\lim_{x \to 0} \frac{e^{2x} - 1}{\sin(3x)}$$

First, substitute $x = 0$ into the numerator: $$e^{2(0)} - 1 = e^0 - 1 = 1 - 1 = 0.$$

Now substitute $x = 0$ into the denominator: $$\sin(3 \cdot 0) = \sin(0) = 0.$$

So the expression has the indeterminate form $$\frac{0}{0}.$$

Since both the numerator and denominator approach $0$, we can use L’Hopital’s Rule.

Differentiate the numerator: $$\frac{d}{dx}\left(e^{2x} - 1\right) = 2e^{2x}.$$

Differentiate the denominator: $$\frac{d}{dx}\left(\sin(3x)\right) = 3\cos(3x).$$

Therefore, $$\lim_{x \to 0} \frac{e^{2x} - 1}{\sin(3x)} = \lim_{x \to 0} \frac{2e^{2x}}{3\cos(3x)}.$$

Now substitute $x = 0$: $$\frac{2e^{2(0)}}{3\cos(3 \cdot 0)} = \frac{2e^0}{3\cos(0)}.$$

Since $$e^0 = 1$$ and $$\cos(0) = 1,$$ we get $$\frac{2e^0}{3\cos(0)} = \frac{2(1)}{3(1)} = \frac{2}{3}.$$

Thus, $$\boxed{ \lim_{x \to 0} \frac{e^{2x} - 1}{\sin(3x)} = \frac{2}{3} }$$