Question
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}.$
✓
Answer
We have, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ax}}{{b\sin x}} + \frac{{x\cos x}}{{b\sin x}}} \right)$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} + \frac{1}{b}\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x}}{{\sin x}}$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\mathop {\lim }\limits_{bx \to 0} \frac{{\cos x}}{{\left( {\frac{{\sin x}}{x}} \right)}}$
$ = \frac{a}{b}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\frac{{\mathop {\lim }\limits_{x \to 0} \cos x}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}}$$\left[\because {\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$
$ = \frac{a}{b} \times \frac{1}{1} + \frac{1}{b} \times \frac{1}{1}$$\left[\because {\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
$= \frac { a + 1 } { b }$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} + \frac{1}{b}\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x}}{{\sin x}}$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\mathop {\lim }\limits_{bx \to 0} \frac{{\cos x}}{{\left( {\frac{{\sin x}}{x}} \right)}}$
$ = \frac{a}{b}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\frac{{\mathop {\lim }\limits_{x \to 0} \cos x}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}}$$\left[\because {\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$
$ = \frac{a}{b} \times \frac{1}{1} + \frac{1}{b} \times \frac{1}{1}$$\left[\because {\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
$= \frac { a + 1 } { b }$
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