Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{ - \sqrt {25 - {x^2}} - \left( { - \sqrt {24} } \right)}}{{x - 1}} \cr & = \mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{\sqrt {24} - \sqrt {25 - {x^2}} }}{{x - 1}} \times \frac{{\sqrt {24} + \sqrt {25 - {x^2}} }}{{\sqrt {24} + \sqrt {25 - {x^2}} }} \cr & = \mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{{x^2} - 1}}{{\left( {x - 1} \right)\left[ {\sqrt {24} + \sqrt {25 - {x^2}} } \right]}} \cr & = \mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{x + 1}}{{\left[ {\sqrt {24} + \sqrt {25 - {x^2}} } \right]}}\,\,\, = \frac{2}{{2\sqrt {24} }}\,\, = \frac{1}{{2\sqrt 6 }} \cr} $$

$$\eqalign{ & {x^2} + 4x + 5 = {\left( {x + 2} \right)^2} + 1 \geqslant 1 \cr & {\text{So, }}a = 1 \cr & b = \mathop {\lim }\limits_{\theta \to 0} \frac{{2\,{{\sin }^2}\theta }}{{{\theta ^2}}} = 2 \cr & \sum\limits_{r = 0}^n {{a^r} \cdot {b^{n - r}}} = {b^n} + a{b^{n - 1}} + {a^2}{b^{n - 2}} + ..... + {a^n} \cr & = \frac{{{b^n}\left[ {1 - {{\left( {\frac{a}{b}} \right)}^{n + 1}}} \right]}}{{1 - \frac{a}{b}}} \cr & = \frac{{{2^n}\left[ {1 - {{\left( {\frac{1}{2}} \right)}^{n + 1}}} \right]}}{{1 - \frac{1}{2}}} \cr & = \frac{{{2^{n + 1}}\left( {{2^{n + 1}} - 1} \right)}}{{{2^{n + 1}}}} \cr & = \left( {{2^{n + 1}} - 1} \right) \cr} $$

$$\eqalign{ & {\text{We have, }}f\left( x \right) = \alpha \left( x \right)\beta \left( x \right)\gamma \left( x \right){\text{, for all real }}x \cr & \Rightarrow f'\left( x \right) = \alpha '\left( x \right)\beta \left( x \right)\gamma \left( x \right) + \alpha \left( x \right)\beta '\left( x \right)\gamma \left( x \right) + \alpha \left( x \right)\beta \left( x \right)\gamma '\left( x \right) \cr & \Rightarrow f'\left( 2 \right) = \alpha '\left( 2 \right)\beta \left( 2 \right)\gamma \left( 2 \right){\text{ + }}\alpha \left( 2 \right)\beta '\left( 2 \right)\gamma \left( 2 \right){\text{ + }}\alpha \left( 2 \right)\beta \left( 2 \right)\gamma '\left( 2 \right) \cr & \Rightarrow 18f\left( 2 \right) = 3\alpha \left( 2 \right)\beta \left( 2 \right)\gamma \left( 2 \right) - 4\alpha \left( 2 \right)\beta \left( 2 \right)\gamma \left( 2 \right) + k\alpha \left( 2 \right)\beta \left( 2 \right)\gamma \left( 2 \right) \cr & \left[ {\because \,f'\left( 2 \right) = 18f\left( 2 \right),\,\alpha '\left( 2 \right) = 3\alpha \left( 2 \right),\,\beta '\left( 2 \right) = - 4\beta \left( 2 \right){\text{ and }}\gamma '\left( 2 \right) = k\gamma \left( 2 \right)} \right] \cr & \Rightarrow 18f\left( 2 \right) = \left( { - 1 + k} \right)\alpha \left( 2 \right)\beta \left( 2 \right)\gamma \left( 2 \right) \cr & \Rightarrow 18f\left( 2 \right) = \left( {k - 1} \right)f\left( 2 \right)\,\,\,\,\,\left[ {\because \,f\left( 2 \right) = \alpha \left( 2 \right)\beta \left( 2 \right)\gamma \left( 2 \right)} \right] \cr & \Rightarrow k - 1 = 18 \cr & \Rightarrow k = 19 \cr} $$

The given expression can be written as
$$\eqalign{ & = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{n}\,\sum\limits_{r\, = \,1}^n {{{\left( {\frac{r}{n}} \right)}^4} - } \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{n}.\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}{\left( {\frac{r}{n}} \right)^3} \cr & = \int\limits_0^1 {{x^4}\,dx - \mathop {\lim }\limits_{n \to \infty } } \frac{1}{n} \times \int\limits_0^1 {{x^3}} \,dx \cr & = \left[ {\frac{{{x^5}}}{5}} \right]_0^1 - 0 \cr & = \frac{1}{5} \cr} $$

$$\eqalign{ & nx - 1 < \left[ {nx} \right] \leqslant nx. \cr & {\text{Putting }}n = 1,\,2,\,3,.....,\,n\,{\text{and adding them,}} \cr & x\sum n - n < \sum {\left[ {nx} \right] \leqslant x\sum n } \cr & \therefore \,x.\frac{{\sum n }}{{{n^2}}} - \frac{1}{n} < \frac{{\sum {\left[ {nx} \right]} }}{{{n^2}}} \leqslant x.\frac{{\sum n }}{{{n^2}}}.....({\text{i}}) \cr & {\text{Now, }}\mathop {\lim }\limits_{n \to \infty } \left\{ {x.\frac{{\sum n }}{{{n^2}}} - \frac{1}{n}} \right\} = x.\mathop {\lim }\limits_{n \to \infty } \frac{{\sum n }}{{{n^2}}} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = \frac{x}{2} \cr & {\text{As the two limits are equal, by }}\left( {\text{i}} \right) \cr & \mathop {\lim }\limits_{n \to \infty } \frac{{\sum {\left[ {nx} \right]} }}{{{n^2}}} = \frac{x}{2} \cr & {\text{ }} \cr} $$

$$\eqalign{ & {\text{Given }}3 = \mathop {\lim }\limits_{x \to 0} {\left( {1 + a\,\sin \,x} \right)^{{\text{cosec}}\,x}}\,\,\left[ {{1^\infty }{\text{ form}}} \right] \cr & {\text{Put sin }}x = 4\,\,\,\,\therefore \,{\text{when}}\,x \to 0,\,y \to 0 \cr & \therefore \,\mathop {\lim }\limits_{x \to 0} {\left( {1 + a\,\sin \,x} \right)^{{\text{cosec}}\,x}} = \mathop {\lim }\limits_{y \to 0} {\left( {1 + ay} \right)^{\frac{1}{y}}} = {e^a} \cr & \therefore \,{e^a} = 3\,\,\, \Rightarrow a = {\log _e}3 = {\text{ln}}\,{\text{3}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \,kx\,{\text{cosec}}\,x = \mathop {\lim }\limits_{x \to 0} \,x\,{\text{cosec}}\,kx \cr & \Rightarrow k.\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin \,x}} = \frac{1}{k}\mathop {\lim }\limits_{x \to 0} \frac{{kx}}{{\sin \,kx}} \cr & \Rightarrow k = \frac{1}{k} \cr & \Rightarrow k = \pm 1 \cr} $$

$$\eqalign{ & {\text{We have,}} \cr & \mathop {\lim }\limits_{x \to \infty } \,{2^{x - 1}}\tan \left( {\frac{a}{{{2^x}}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } \,\frac{a}{2}\frac{{\tan \left( {\frac{a}{{{2^x}}}} \right)}}{{\left( {\frac{a}{{{2^x}}}} \right)}}\,\,\,\left( {\frac{0}{0}{\text{ form}}} \right) \cr & = \frac{a}{2}\mathop {\lim }\limits_{y \to 0} \frac{{\tan \,y}}{y} = \frac{a}{2}\,\,\,\,\,\left( {{\text{where }}y = \frac{a}{{{2^x}}}} \right) \cr} $$

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

$$\eqalign{ & {t_r} = \frac{1}{2}.\frac{{2r}}{{{{\left( {{r^2} + 1} \right)}^2} - {r^2}}} \cr & = \frac{1}{2}\left\{ {\frac{1}{{{r^2} - r + 1}} - \frac{1}{{{r^2} + r + 1}}} \right\} \cr & = \frac{1}{2}\left\{ {\frac{1}{{r\left( {r - 1} \right) + 1}} - \frac{1}{{\left( {r + 1} \right)r + 1}}} \right\} \cr & \therefore \sum\limits_{r = 1}^n {{t_r}} = \sum\limits_{r = 1}^n {\frac{1}{2}\left\{ {f\left( r \right) - f\left( {r + 1} \right)} \right\}} {\text{, where }}f\left( r \right) = \frac{1}{{r\left( {r - 1} \right) + 1}} \cr & = \frac{1}{2}\left\{ {f\left( 1 \right) - f\left( {n + 1} \right)} \right\} \cr & = \frac{1}{2}\left\{ {1 - \frac{1}{{\left( {n + 1} \right)n + 1}}} \right\} \to \frac{1}{2}{\text{ as }}n \to \infty \cr} $$