Question

Evaluate the limit:

$\lim_{x \to 0} \frac{\sin(3x)}{2x}$

Answer 1

3/2

Answer 2

To evaluate the limit $\lim_{x \to 0} \frac{\sin(3x)}{2x}$, we can use the fundamental trigonometric limit: $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$

The given limit is: $\lim_{x \to 0} \frac{\sin(3x)}{2x}$

As $x \to 0$, both the numerator $\sin(3x)$ and the denominator $2x$ approach $0$, resulting in an indeterminate form $\frac{0}{0}$.

To apply the standard limit formula, we need the argument of the sine function to be present in the denominator. The argument is $3x$. So, we need $3x$ in the denominator. We can achieve this by multiplying and dividing by $3$:

$\lim_{x \to 0} \frac{\sin(3x)}{2x} = \lim_{x \to 0} \frac{\sin(3x)}{2x} \times \frac{3}{3}$

Rearrange the terms to group $\frac{\sin(3x)}{3x}$: $= \lim_{x \to 0} \frac{\sin(3x)}{3x} \times \frac{3}{2}$

Now, we can separate the limit into two parts: $= \left( \lim_{x \to 0} \frac{\sin(3x)}{3x} \right) \times \left( \lim_{x \to 0} \frac{3}{2} \right)$

For the first part, let $y = 3x$. As $x \to 0$, $y \to 3(0) = 0$. So, the first limit becomes: $\lim_{y \to 0} \frac{\sin(y)}{y} = 1$

The second part is a constant, so its limit is the constant itself: $\lim_{x \to 0} \frac{3}{2} = \frac{3}{2}$

Substitute these values back into the expression: $= 1 \times \frac{3}{2} = \frac{3}{2}$

Thus, the value of the limit is $\frac{3}{2}$.