The value of \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{\sin x}}} \right)\)

Concept:

For limit evaluation of indeterminate forms, i.e. for \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\) form, we apply L’ Hospital’s Rule as:

\(\mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right\} \)

Calculation:

The given limit is,

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sinx}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x ~- ~x}}{{xsin\;x}}} \right) \)

By putting a limit we get \(\frac{0}{0}\) form, therefore, by L'Hospital Rule:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sin~x}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos x~ - ~1}}{{xcos\;x ~+ ~\sin x}}} \right) \)

Again putting the value of limit we get \(\frac{0}{0}\) form, therefore, again using L'Hospital Rule:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sinx}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin x}}{{x\left( { - \sin x} \right)~ +~ \cos x ~+~ \cos x}}} \right) \)

\( = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin x}}{{2\cos x -~~ xsin\;x}}} \right) \)

Now, by putting the limit, we get:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{\sin x}}} \right) = \frac{0}{{2 - 0}} = 0 \)

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sin~x}}} \right)~=~0 \)