Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = \sqrt {\frac{{x - \sin \,x}}{{x + {{\cos }^2}x}}} \,; \cr & \Rightarrow \mathop {\lim }\limits_{x\, \to \,\infty } f\left( x \right) = \mathop {\lim }\limits_{x\, \to \,\infty } \sqrt {\frac{{1 - \frac{{\sin \,x}}{x}}}{{1 + \frac{{{{\cos }^2}\,x}}{x}}}} = \sqrt {\frac{{1 - 0}}{{1 + 0}}} = 1 \cr} $$

$${\text{Limit}} = {e^{\mathop {\lim }\limits_{x \to 0} \,\frac{1}{x}\,\log \left\{ {\tan \left( {\frac{\pi }{4} - x} \right)} \right\}}} = {e^{{{\mathop {\lim }\limits_{x \to 0} }^{\frac{{\cot \left( {\frac{\pi }{4} - x} \right).{{\sec }^2}\left( {\frac{\pi }{4} - x} \right).\left( { - 1} \right)}}{1}}}}} = {e^{ - 2}}$$

$$\eqalign{ & {\text{Area of }}\Delta PQR = A \cr & = \frac{{ - 1}}{2}\left[ {x\left( {b - 0} \right) + 0\left( {0 - y} \right) + a\left( {y - b} \right)} \right] \cr & = \frac{{ - 1}}{2}\left( {bx + ax - ab} \right) \cr & \therefore \,\frac{{dA}}{{dx}} = \frac{{ - 1}}{2}\left( {a + b} \right) \cr} $$

As $$\alpha $$ is a repeated root, $$a{\alpha ^2} + b\alpha + c = 0$$    and $$2a\alpha + b = 0$$
$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to \alpha } \frac{{\left( {2ax + b} \right)\cos \left( {a{x^2} + bx + c} \right)}}{{2\left( {x - \alpha } \right)}} \cr & = \mathop {\lim }\limits_{x \to \alpha } \frac{{2a\,\cos \left( {a{x^2} + bx + c} \right) - {{\left( {2ax + b} \right)}^2}\sin \left( {a{x^2} + bx + c} \right)}}{2} \cr & = \frac{{2a}}{2} \cr & = a \cr} $$

Given limit can be written as
$$\eqalign{ & \mathop {\lim }\limits_{n \to 0} \frac{{{{\sin }^2}\left( {n!} \right)}}{{{n^{1 - p}}\left( {1 + \frac{1}{n}} \right)}}\,\,\left( {0 < p < 1} \right) \cr & = \frac{{{\text{some real number in }}\left[ {0,\,1} \right]}}{\infty } \cr & = 0\,\,\,\left( {\because \,1 - p > 0} \right) \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{1 + {x^2}}}}}{{2x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{1 + {x^2} - \sqrt {1 - {x^2}} }}{{2x\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 + {x^2}} \right)}^2} - \left( {1 - {x^2}} \right)}}{{2x\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} \left\{ {1 + {x^2} + \sqrt {1 - {x^2}} } \right\}}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{x^4} + 3{x^2}}}{{2x\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} \left\{ {1 + {x^2} + \sqrt {1 - {x^2}} } \right\}}} \cr & = 0 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sin \,{x^4} - {x^4}\cos \,{x^4} + {x^{20}}}}{{{x^4}\left( {{e^{2{x^4}}}1 - 2{x^4}} \right)}} \cr & = \mathop {\lim }\limits_{t \to 0} \frac{{\sin \,t - t\,\cos \,t + {t^5}}}{{t\left( {{e^{2t}} - 1 - 2t} \right)}} \cr & = \mathop {\lim }\limits_{t \to 0} \frac{{t - \frac{{{t^3}}}{{3!}} + \frac{{{t^5}}}{{5!}}..... - t\left( {1 - \frac{{{t^2}}}{{2!}} + \frac{{{t^4}}}{{4!}}.....} \right) + {t^5}}}{{t\left( {1 + 2t + \frac{{4{t^2}}}{{2!}} + \frac{{8{t^3}}}{{3!}} + \frac{{16{t^4}}}{{4!}} + ..... - 1 - 2t} \right)}} \cr & = \mathop {\lim }\limits_{t \to 0} \frac{{ - \frac{{{t^3}}}{6} + \frac{{{t^3}}}{2} + \frac{{{t^5}}}{{5!}} - \frac{{{t^5}}}{{4!}} + ..... + {t^5}}}{{2{t^3} + \frac{{8{t^4}}}{{3!}} + .....}} \cr & = \frac{{ - \frac{1}{6} + \frac{1}{2}}}{2} \cr & = \frac{{ - 1 + 3}}{{12}} \cr & = \frac{1}{6} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} }}{{x - 2}} \cr & = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt 2 \left| {\sin \left( {x - 2} \right)} \right|}}{{x - 2}} \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\sqrt 2 \sin \left( {x - 2} \right)}}{{\left( {x - 2} \right)}} = - \sqrt 2 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt 2 \sin \left( {x - 2} \right)}}{{\left( {x - 2} \right)}} = \sqrt 2 \cr} $$
Thus L.H.L. $$ \ne $$ R.H.L.
Hence, $$\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} }}{{x - 2}}$$     does not exist

$$\eqalign{ & {\text{For }}\left| x \right| < 1,\,{x^{2n}} \to 0{\text{ as }}n \to \infty {\text{ and}} \cr & {\text{For }}\left| x \right| > 1,\,\frac{1}{{{x^{2n}}}} \to 0{\text{ as }}n \to \infty \cr & {\text{So,}} \cr} $$
\[f\left( x \right) = \left\{ \begin{array}{l} \log \left( {2 + x} \right){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}\left| x \right| < 1\\ \mathop {\lim }\limits_{n \to \infty } \frac{{{x^{ - 2n}}\log \left( {2 + x} \right) - \sin \,x}}{{{x^{ - 2n}} + 1}} = - \sin \,x\,\,\,\,\,\,{\rm{if }}\left| x \right| > 1\\ \frac{1}{2}\left[ {\log \left( {2 + x} \right) - \sin \,x} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}\left| x \right| = 1 \end{array} \right.\]
$$\eqalign{ & {\text{Thus }}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x{ \to ^{1 + }}} \left( { - \sin \,x} \right) = - \sin \,1 \cr & {\text{and }}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x{ \to ^{1 - }}} \,\log \left( {2 + x} \right) = \log \,3 \cr & {\text{Hence, L}}{\text{.H}}{\text{.L. }} \ne {\text{ R}}{\text{.H}}{\text{.L}}{\text{.}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {\log _e}{\left( {\sin \,x} \right)^{\tan \,x}} \cr & = \mathop {\lim }\limits_{x \to 0} \tan \,x \cdot {\log _e}\sin \,x\,\,\,\,\left( {0 \cdot \infty \,{\text{form}}} \right) \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}\sin \,x}}{{\cot \,x}}\,\,\,\,\left( {\frac{\infty }{\infty }{\text{ form}}} \right) \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\cot \,x}}{{ - {\text{cose}}{{\text{c}}^2}x}}\,\,\,\,\left[ {{\text{Using L'Hospital's rule}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \left( { - \cos \,x \cdot \sin \,x} \right) \cr & = 0 \cr} $$