Question

35-38: Use continuity to evaluate the limit.

36. \(\mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \sin \left( {\tan \left( {\cos \theta } \right)} \right)\)

Short answer

The value of the limit is 0.

Step-by-step solution

Theorems of continuity

When the functions say \(f\) and \(g\) are continuous at some number. Then, the following functions are also continuous:

1. \(f + g\)
2. \(f - g\)
3. \(cf\)
4. \(fg\)
5. \(\frac{f}{g},\,\,{\mathop{\rm if}\nolimits} \,\,g\left( a \right) \ne 0\)

Eachpolynomial iscontinuous on\(\mathbb{R} = \left( { - \infty ,\infty } \right)\). Anyrational function is continuous in its domain.

When\(f\), and\(g\)are continuous at some number, then thecomposite function \(f \circ g\)is \(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\) also continuous.

Use continuity to evaluate the limit

It is observed that the function\(f\left( \theta \right) = \sin \left( {\tan \left( {\cos \theta } \right)} \right)\)is a composite of a trigonometric function.

Therefore, the function\(f\left( \theta \right)\)is continuous throughout its domain.

It follows that the domain of\(\tan \theta \)is\(\left( { - 1,1} \right)\)and domain of\(\sin \theta \)is\(\mathbb{R}\), therefore, the domain of the function\(f\)is\(\mathbb{R}\).

Hence, the function\(f\)is continuous at\(\frac{\pi }{2}\).

Evaluate the limit as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \sin \left( {\tan \left( {\cos \frac{\pi }{2}} \right)} \right) &= \sin \left( {\tan 0} \right)\\ &= \sin \left( 0 \right)\\ &= 0\end{aligned}\)

Thus, the value of the limit is 0.