Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \min \left( {{y^2} - 4y + 11} \right) = \min \left[ {{{\left( {y - 2} \right)}^2} + 7} \right] = 7 \cr & {\text{or }}L = \mathop {\lim }\limits_{x \to 0} \left[ {\min \left( {{y^2} - 4y + 11} \right)\frac{{\sin \,x}}{x}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{7\,\sin \,x}}{x}} \right] \cr & = \left[ {{\text{a value slightly lesser than 7}}} \right]\left( {\left| {\sin \,x} \right| < \left| x \right|,\,{\text{when }}x \to 0} \right) \cr & = \mathop {\lim }\limits_{x \to 0} \left[ {7\frac{{\sin \,x}}{x}} \right] \cr & = 6 \cr} $$

$$\eqalign{ & {\text{We have for }}x \in {\bf{R}}, \cr & x - 1 < \left[ x \right] \leqslant x \cr & {\text{or, }}nx - 1 < \left[ {nx} \right] \leqslant nx\left[ {{\text{For }}n \in {\bf{N}}} \right] \cr & {\text{Then,}} \cr & \left( {x - 1} \right) + \left( {2x - 1} \right) + ... + \left( {nx - 1} \right) < \left[ x \right] + \left[ {2x} \right] + ... + \left[ {nx} \right] \leqslant x + 2x + ... + nx \cr & \Rightarrow \frac{{n\left( {n + 1} \right)}}{2}x - n < \left[ x \right] + \left[ {2x} \right] + ... + \left[ {nx} \right] \leqslant \frac{{n\left( {n + 1} \right)}}{2}x \cr & \Rightarrow \left( {\frac{1}{2} + \frac{1}{{2n}}} \right)x - \frac{1}{n} < \frac{{\left[ x \right] + \left[ {2x} \right] + ... + \left[ {nx} \right]}}{{{n^2}}} \leqslant \left( {\frac{1}{{2n}} + \frac{1}{{2n}}} \right)x \cr & \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{2} + \frac{1}{{2n}}} \right)x - \frac{1}{n} \leqslant \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ x \right] + \left[ {2x} \right] + ... + \left[ {nx} \right]}}{{{n^2}}} \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{2} + \frac{1}{{2n}}} \right)x \cr & \left[ {{\text{Using limit property}}} \right] \cr & \Rightarrow \frac{x}{2} \leqslant \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ x \right] + \left[ {2x} \right] + ... + \left[ {nx} \right]}}{{{n^2}}} \leqslant \frac{x}{2} \cr & {\text{Using Sandwich property we get,}} \cr & \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ x \right] + \left[ {2x} \right] + ... + \left[ {nx} \right]}}{{{n^2}}} = \frac{x}{2} \cr} $$

$${\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^{\frac{3}{2}}}\sqrt {2y - x} }}{{{x^{\frac{3}{2}}}\left( {\sqrt {8y - 4x} + \sqrt {8y} } \right)}} = \frac{{\sqrt {2y} }}{{2\sqrt {8y} }} = \frac{1}{4}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\tan \,2x - n\,\sin \,x}}{{{x^3}}} = I \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2x + \frac{{8{x^3}}}{{3!}}..... - nx + \frac{{n{x^3}}}{{3!}}}}{{{x^3}}} = I \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - n} \right)x + \left( {\frac{{16 + n}}{6}} \right){x^3} + .....}}{{{x^3}}} = I \cr & = n = 2{\text{ and, thus required value }} = \frac{{16 + n}}{6} = 3 \cr} $$

Given:
$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4 \cr & \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + x + 1 - a{x^2} - ax - bx - b}}{{x + 1}} = 4 \cr & \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 - a} \right){x^2} + \left( {1 - a - b} \right)x + \left( {1 - b} \right)}}{{x + 1}} = 4 \cr} $$
For this limit to be finite $$1 - a = 0\,\,\,\, \Rightarrow a = 1$$
Then given limit reduces to
$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{ - bx + \left( {1 - b} \right)}}{{x + 1}} = 4 \cr & \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{ - b + \frac{{\left( {1 - b} \right)}}{x}}}{{1 + \frac{1}{x}}} = 4 \cr & \Rightarrow - b = 4\,\,\,\,{\text{or}}\,\,\,b = - 4 \cr & {\text{Hence}}\,\,\,a = 1,\,\,\,b = - 4 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \mathop \prod \limits_{r = 3}^n \frac{{{r^3} - 8}}{{{r^3} + 8}} \cr & = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{3^3} - 8}}{{{3^3} + 8}}} \right)\left( {\frac{{{4^3} - 8}}{{{4^3} + 8}}} \right).....\left( {\frac{{{n^3} - 8}}{{{n^3} + 8}}} \right) \cr & = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{3 - 2}}{{3 + 2}}.\frac{{{3^2} + 4 + 2\left( 3 \right)}}{{{3^2} + 4 - 2\left( 3 \right)}}} \right)\left( {\frac{{4 - 2}}{{4 + 2}}.\frac{{{4^2} + 4 + 2\left( 4 \right)}}{{{4^2} + 4 - 2\left( 4 \right)}}} \right).....\left( {\frac{{n - 2}}{{n + 2}}.\frac{{{n^2} + 4 + 2n}}{{{n^2} + 4 - 2n}}} \right) \cr & = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{3 - 2}}{{3 + 2}}.\frac{{4 - 2}}{{4 + 2}}.\frac{{5 - 2}}{{5 + 2}}.....\frac{{n - 2}}{{n + 2}}} \right)\left( {\frac{{{3^2} + 4 + 2\left( 3 \right)}}{{{3^2} + 4 - 2\left( 3 \right)}}.\frac{{{4^2} + 4 + 2\left( 4 \right)}}{{{4^2} + 4 - 2\left( 4 \right)}}.....\frac{{{n^2} + 4 + 2n}}{{{n^2} + 4 - 2n}}} \right) \cr & = \left( {\frac{{1.2.3.4.5.6.7.....}}{{5.6.7.8.....}}} \right)\left( {\frac{{19.28.39.52.63.....}}{{7.12.19.28.39.52.....}}} \right) \cr & = \frac{{1.2.3.4}}{{7.12}} \cr & = \frac{2}{7} \cr} $$

$$\eqalign{ & {\text{When }}x \to 0,\,\left\{ x \right\} \to 0 \cr & \therefore {\text{Limit}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{\tan \,h}}{\text{ or }}\mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{\tan \,\left( { - h} \right)}} = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}} \cr & \mathop {\lim }\limits_{x \to 2} \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}} = L \cr} $$
It is $$\frac{0}{0}$$ (undefined) condition so using L'hospital's rule
$$\eqalign{ & \Rightarrow L = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{\left\{ {\sin \left( {{e^{x - 2}} - 1} \right)} \right\}}}{{\left\{ {\ln \left( {x - 1} \right)} \right\}}}} \right] \cr & \Rightarrow L = \mathop {\lim }\limits_{x \to 2} \frac{{\cos \left( {{e^{x - 2}} - 1} \right).{e^{\left( {x - 2} \right)}}}}{{\frac{1}{{\left( {x - 1} \right)}}}} \cr & \Rightarrow L = \mathop {\lim }\limits_{x \to 2} \cos \left( {{e^{2 - 2}} - 1} \right){e^{2 - 2}}.\left( {2 - 1} \right) \cr & \Rightarrow L = \cos \left( 0 \right){e^0}.1 \cr & \Rightarrow L = 1 \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \,x}}{{{x^2}}}.\frac{x}{{{2^x} - 1}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2\,{{\sin }^2}\frac{x}{2}}}{{4.{{\left( {\frac{x}{2}} \right)}^2}}}.\frac{x}{{{e^{x\,{{\log }_e}2}} - 1}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{1}{2}.{\left( {\frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2}.\frac{x}{{\left\{ {1 + x\,{{\log }_e}2 + \frac{{{{\left( {x\,{{\log }_e}2} \right)}^2}}}{{n!}} + .....} \right\} - 1}} \cr & = \frac{1}{2}.\frac{1}{{{{\log }_e}2}} \cr & = \frac{1}{2}{\log _2}e\, \cr} $$

$$\eqalign{ & {\text{Given }}\mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} {\left[ {1 + {{\left( {\cos \,x} \right)}^{\cos \,x}}} \right]^2} \cr & {\text{Let }}y = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} {\left( {\cos \,x} \right)^{\cos \,x}} \cr & \log \left( y \right) = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \left( {\cos \,x} \right)\log \,\cos \,x \cr & \log \left( y \right) = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \frac{{\log \left( {\cos \,x} \right)}}{{\sec \,\left( x \right)}}\,\,\,\left( {\frac{\infty }{\infty }{\text{ form}}} \right) \cr & {\text{Applying L'Hospital's rule}} \cr & \log \left( y \right) = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \frac{{ - \sin \,x}}{{\cos \,x\left( {\sec \,x\,\tan \,x} \right)}} \cr & = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \left( { - \cos \,x} \right) \cr & = 0 \cr & \therefore \,\,y = {e^0} = 1 \cr & {\text{Now, limit is}}\,{\left( {1 + 1} \right)^2} = {2^2} = 4 \cr} $$