Question

$$\mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {{y^2} - {{\left( {y - x} \right)}^2}} }}{{{{\left( {\sqrt {8xy - 4{x^2}} + \sqrt {8xy} } \right)}^3}}}$$      is equal to :

A. $$\frac{1}{4}$$  
B. $$\frac{1}{2}$$
C. $$\frac{1}{{2\sqrt 2 }}$$
D. none of these
Answer :   $$\frac{1}{4}$$
Solution :
$${\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^{\frac{3}{2}}}\sqrt {2y - x} }}{{{x^{\frac{3}{2}}}\left( {\sqrt {8y - 4x} + \sqrt {8y} } \right)}} = \frac{{\sqrt {2y} }}{{2\sqrt {8y} }} = \frac{1}{4}$$

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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