Question

1. - / 1 points find the limit for the given function. $lim_{t \to 0} \frac{(sin(2t))^{2}}{t^{2}}$

Explanation:

Step1: Rewrite the function

We know that $\lim_{t
ightarrow0}\frac{\sin(2t)}{2t} = 1$. First, rewrite $\frac{(\sin(2t))^{2}}{t^{2}}$ as $4\times\frac{(\sin(2t))^{2}}{(2t)^{2}}$.

Step2: Apply the limit rule

$\lim_{t
ightarrow0}\frac{(\sin(2t))^{2}}{t^{2}}=\lim_{t
ightarrow0}4\times\frac{(\sin(2t))^{2}}{(2t)^{2}}$. According to the constant - multiple rule of limits $\lim_{t
ightarrow a}(cf(t)) = c\lim_{t
ightarrow a}f(t)$ and the fact that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$ (let $u = 2t$, as $t
ightarrow0$, $u
ightarrow0$), we have $\lim_{t
ightarrow0}4\times\frac{(\sin(2t))^{2}}{(2t)^{2}}=4\lim_{t
ightarrow0}(\frac{\sin(2t)}{2t})^{2}$.

Step3: Calculate the limit

Since $\lim_{t
ightarrow0}\frac{\sin(2t)}{2t}=1$, then $4\lim_{t
ightarrow0}(\frac{\sin(2t)}{2t})^{2}=4\times1^{2}=4$.

Answer:

$4$