Question

use the relation \\(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\\) to determine the limit. \\(\lim_{t \to 0} \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 \to 0} \frac{\sin kt}{t} = \square\\) b. the limit does not ex...

Explanation:

Step1: Rewrite the expression

We have \(\lim_{t \to 0} \frac{\sin kt}{t}\). Multiply and divide by \(k\) (where \(k
eq0\), and if \(k = 0\), the limit is trivially \(0\)):
\(\lim_{t \to 0} \frac{\sin kt}{t}=\lim_{t \to 0} k\cdot\frac{\sin kt}{kt}\)

Step2: Apply the limit formula

Let \(\theta=kt\). As \(t\to0\), \(\theta\to0\). We know that \(\lim_{\theta\to0}\frac{\sin\theta}{\theta} = 1\). So:
\(k\cdot\lim_{kt \to 0} \frac{\sin kt}{kt}=k\cdot1\) (since \(\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1\) with \(\theta = kt\))

Answer:

For option A, \(\lim_{t \to 0} \frac{\sin kt}{t}=k\)