Question
A function $$f\left( x \right)$$ is defined as below
$$f\left( x \right) = \frac{{\cos \left( {\sin \,x} \right) - \cos \,x}}{{{x^2}}},\,x \ne 0{\text{ and }}f\left( 0 \right) = a$$
$$f\left( x \right)$$ is continuous at $$x=0$$ if $$a$$ equals :
A.
$$0$$
B.
$$4$$
C.
$$5$$
D.
6
Answer :
$$0$$
Solution :
$$\eqalign{
& {\text{RH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left\{ {\sin \left( {0 + h} \right)} \right\} - \cos \left( {0 + h} \right)}}{{{{\left( {0 + h} \right)}^2}}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \,\sin \,h - \cos \,h}}{{{h^2}}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {\sin \,h} \right) \times \cos \,h + \sin \,h}}{{2h}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{ - \cos \left( {\sin \,h} \right) \times {{\cos }^2}h + \sin \left( {\sin \,h} \right) \times \sin \,h + \cos \,h}}{2} \cr
& = 0 \cr} $$
Similarly, LH limit $$=0.$$
As $$f\left( x \right)$$ is continuous at $$x = 0,\,f\left( 0 \right) = 0$$