Question

find the following limit or state that it does not exist.
\\(\lim \limits_{x \to \pi} \frac{\sin 2x}{2 \sin x}\\)

select the correct choice and, if necessary, fill in the answer box to complete your choice.
\\(\bigcirc\\) a. \\(\lim \limits_{x \to \pi} \frac{\sin 2x}{2 \sin x} = \square\\) (simplify your answer.)
\\(\bigcirc\\) b. the limit does not exist

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula for sine: $\sin2\alpha = 2\sin\alpha\cos\alpha$. For $\alpha=x$, we have $\sin2x = 2\sin x\cos x$.
Substitute $\sin2x = 2\sin x\cos x$ into the function $\frac{\sin2x}{2\sin x}$:
$\frac{\sin2x}{2\sin x}=\frac{2\sin x\cos x}{2\sin x}$

Step2: Simplify the function (considering $\sin x

eq0$ near $x = \pi$)
We can cancel out the common factor $2\sin x$ (since when $x
ightarrow\pi$, $\sin x=\sin\pi = 0$, but in the neighborhood of $x = \pi$ (excluding $x=\pi$), $\sin x
eq0$), so $\frac{2\sin x\cos x}{2\sin x}=\cos x$ (for $x
eq k\pi,k\in\mathbb{Z}$ and $x$ near $\pi$).

Step3: Evaluate the limit

Now we find the limit of $\cos x$ as $x
ightarrow\pi$. We know that the cosine function is continuous everywhere, so $\lim_{x
ightarrow\pi}\cos x=\cos\pi=- 1$.

Answer:

A. $\lim\limits_{x
ightarrow\pi}\frac{\sin2x}{2\sin x}=-1$