Question

find the limit using \\(\lim\limits_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1\\).
\\(\lim\limits_{x \to 0} \dfrac{x \csc 10x}{\cos 8x}\\)

select the correct choice below and, if necessary, fill in the answer box in your choice.

\\(\bigcirc\\) a. \\(\lim\limits_{x \to 0} \dfrac{x \csc 10x}{\cos 8x} = \square\\) (simplify your answer.)
\\(\bigcirc\\) b. the limit does not exist.

Explanation:

Step1: Recall csc definition

Recall that \(\csc\theta=\frac{1}{\sin\theta}\), so rewrite the numerator:
\(x\csc 10x = x\cdot\frac{1}{\sin 10x}=\frac{x}{\sin 10x}\)

Step2: Manipulate to use the limit formula

We know \(\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta} = 1\), so \(\lim_{\theta
ightarrow0}\frac{\theta}{\sin\theta}=1\) (reciprocal). Let \(\theta = 10x\), as \(x
ightarrow0\), \(10x
ightarrow0\). Then \(\frac{x}{\sin 10x}=\frac{1}{10}\cdot\frac{10x}{\sin 10x}\)

Step3: Evaluate the limit of the numerator and denominator

First, find \(\lim_{x
ightarrow0}\frac{x}{\sin 10x}=\lim_{x
ightarrow0}\frac{1}{10}\cdot\frac{10x}{\sin 10x}=\frac{1}{10}\cdot\lim_{10x
ightarrow0}\frac{10x}{\sin 10x}=\frac{1}{10}\cdot1=\frac{1}{10}\) (using the limit formula). Then find \(\lim_{x
ightarrow0}\cos 8x=\cos(0) = 1\) (since \(\cos\) is continuous at \(0\))

Step4: Use quotient rule for limits

The limit of a quotient is the quotient of the limits (if the limit of the denominator is non - zero). So \(\lim_{x
ightarrow0}\frac{x\csc 10x}{\cos 8x}=\frac{\lim_{x
ightarrow0}x\csc 10x}{\lim_{x
ightarrow0}\cos 8x}=\frac{\frac{1}{10}}{1}=\frac{1}{10}\)

Answer:

A. \(\lim\limits_{x
ightarrow0}\frac{x\csc 10x}{\cos 8x}=\frac{1}{10}\)