Question

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos \,2x} }}{{\sqrt 2 x}}$$     is-

A. $$1$$
B. $$-1$$
C. zero
D. does not exist  
Answer :   does not exist
Solution :
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos \,2x} }}{{\sqrt 2 x}}\,\, \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}x} \right)} }}{{\sqrt 2 x}}; \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2\,{{\sin }^2}x} }}{{\sqrt 2 x}}\,\,\, \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\left| {\sin \,x} \right|}}{x} \cr} $$
The limit of above does not exist as
$${\text{LHS}} = - 1 \ne \,\,{\text{RHL}} = 1$$

Releted MCQ Question on
Calculus >> Limits

Releted Question 1

lf $$f\left( x \right) = \sqrt {\frac{{x - \sin \,x}}{{x + {{\cos }^2}x}}} ,$$     then $$\mathop {\lim }\limits_{x\, \to \,\infty } f\left( x \right)$$    is-

A. $$0$$
B. $$\infty $$
C. $$1$$
D. none of these
Releted Question 2

If $$G\left( x \right) = - \sqrt {25 - {x^2}} $$     then $$\mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{G\left( x \right) - G\left( I \right)}}{{x - 1}}$$     has the value-

A. $$\frac{1}{{24}}$$
B. $$\frac{1}{{5}}$$
C. $$ - \sqrt {24} $$
D. none of these
Releted Question 3

$$\mathop {\lim }\limits_{n\, \to \,\infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + ..... + \frac{n}{{1 - {n^2}}}} \right\}$$        is equal to-

A. $$0$$
B. $$ - \frac{1}{2}$$
C. $$ \frac{1}{2}$$
D. none of these
Releted Question 4

If $$\eqalign{ & f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ x \right] = 0 \cr} $$
Where \[\left[ x \right]\] denotes the greatest integer less than or equal to $$x.$$ then $$\mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)$$   equals

A. $$1$$
B. $$0$$
C. $$ - 1$$
D. none of these

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Limits


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