Question

question evaluate the limit: \\(\lim\limits_{x\to 0} \frac{-3\sin(2x)}{x}\\) answer attempt 1 out of 2

Explanation:

Step1: Recall the limit formula

We know that $\lim_{u \to 0} \frac{\sin(u)}{u}=1$. We need to adjust our given limit to use this formula. The given limit is $\lim_{x \to 0} \frac{-3\sin(2x)}{x}$. Let's rewrite the denominator and numerator to match the form of the known limit.

Step2: Manipulate the expression

We can rewrite the limit as $\lim_{x \to 0} -3\times 2\times\frac{\sin(2x)}{2x}$. Let $u = 2x$. As $x \to 0$, $u \to 0$. So now our limit becomes $-6\lim_{u \to 0} \frac{\sin(u)}{u}$.

Step3: Apply the limit formula

Using the formula $\lim_{u \to 0} \frac{\sin(u)}{u}=1$, we substitute this into our expression. So $-6\times1=-6$.

Answer:

-6