Question

estimate the limit numerically or state that the limit does not exist (dne):
lim_{x \to 0} \frac{sin(2x)}{x}
give your answer to at least three decimal places
question help: video 1 video 2 message instructor

Explanation:

Step1: Use trigonometric - limit identity

We know the well - known limit $\lim_{u
ightarrow0}\frac{\sin(u)}{u}=1$. Let $u = 2x$. As $x
ightarrow0$, then $u
ightarrow0$. And $\frac{\sin(2x)}{x}=2\times\frac{\sin(2x)}{2x}$.

Step2: Apply the limit

$\lim_{x
ightarrow0}\frac{\sin(2x)}{x}=\lim_{x
ightarrow0}(2\times\frac{\sin(2x)}{2x})$. By the constant - multiple rule for limits $\lim_{x
ightarrow a}(cf(x)) = c\lim_{x
ightarrow a}f(x)$ where $c = 2$ and $f(x)=\frac{\sin(2x)}{2x}$. So $\lim_{x
ightarrow0}(2\times\frac{\sin(2x)}{2x})=2\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}$. Since $\lim_{u
ightarrow0}\frac{\sin(u)}{u}=1$ and $u = 2x$ (as $x
ightarrow0$, $u
ightarrow0$), then $2\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}=2\times1$.

Answer:

$2.000$