Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.
$lim_{t
ightarrow0}\frac{sin kt}{t}$ (k constant)
select the correct answer below and, if necessary, fill in the answer - box to complete your choice.
a. $lim_{t
ightarrow0}\frac{sin kt}{t}=square$
b. the limit does not exist.

Explanation:

Step1: Rewrite the limit

Let $u = kt$. As $t
ightarrow0$, then $u
ightarrow0$. And $t=\frac{u}{k}$ (assuming $k
eq0$). So $\lim_{t
ightarrow0}\frac{\sin kt}{t}=\lim_{u
ightarrow0}\frac{\sin u}{\frac{u}{k}}$.

Step2: Simplify the expression

$\lim_{u
ightarrow0}\frac{\sin u}{\frac{u}{k}}=\lim_{u
ightarrow0}k\cdot\frac{\sin u}{u}$.

Step3: Apply the given limit relation

Since $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$, then $\lim_{u
ightarrow0}k\cdot\frac{\sin u}{u}=k\cdot\lim_{u
ightarrow0}\frac{\sin u}{u}=k\times1 = k$.

Answer:

A. $\lim_{t
ightarrow0}\frac{\sin kt}{t}=k$