Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Let }}\mathop {\lim }\limits_{n \to \infty } \,{b_n} = b \cr & {\text{Now, }}{b_{n + 1}} = \frac{1}{3}\left( {2{b_n} + \frac{{125}}{{b_n^2}}} \right) \cr & {\text{or }}\mathop {\lim }\limits_{n \to \infty } {b_{n + 1}} = \frac{1}{3}\left( {2\mathop {\lim }\limits_{n \to \infty } {b_n} + \frac{{125}}{{\mathop {\lim }\limits_{n \to \infty } b_n^2}}} \right) \cr & {\text{or }}b = \frac{1}{3}\left( {2b + \frac{{125}}{{{b^2}}}} \right) \cr & \Rightarrow \frac{b}{3} = \frac{{125}}{{3{b^2}}} \cr & \Rightarrow {b^3} = 125{\text{ or }}b = 5 \cr} $$

Since $$\mathop {\lim }\limits_{x \to \infty } \left[ x \right]$$  does not exist, hence the required limit does not exist.

$$\eqalign{ & {\text{Consider }}\mathop {\lim }\limits_{x \to 0} {\left\{ {\frac{{1 + \tan \,x}}{{1 + \sin \,x}}} \right\}^{{\text{cosec}}\,x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left[ {{{\left( {1 + \frac{{\sin \,x}}{{\cos \,x}}} \right)}^{\frac{{\cos \,x}}{{\sin \,x}}}}} \right]}^{\frac{1}{{\cos \,x}}}}}}{{{{\left( {1 + \sin \,x} \right)}^{\frac{1}{{\sin \,x}}}}}} \cr & {\text{We know, }}\mathop {\lim }\limits_{n \to 0} {\left( {1 + \frac{1}{n}} \right)^n} = e \cr & \therefore \,\mathop {\lim }\limits_{x \to 0} \frac{{{{\left[ {{{\left( {1 + \frac{{\sin \,x}}{{\cos \,x}}} \right)}^{\frac{{\cos \,x}}{{\sin \,x}}}}} \right]}^{\frac{1}{{\cos \,x}}}}}}{{{{\left( {1 + \sin \,x} \right)}^{\frac{1}{{\sin \,x}}}}}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left[ {{{\left( {1 + \frac{1}{{\frac{{\cos \,x}}{{\sin \,x}}}}} \right)}^{\frac{{\cos \,x}}{{\sin \,x}}}}} \right]}^{\frac{1}{{\cos \,x}}}}}}{{\left[ {{{\left( {1 + \frac{1}{{{\text{cosec}}\,x}}} \right)}^{{\text{cosec}}\,x}}} \right]}} \cr & = \frac{{{e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos \,x}}}}}}{e} \cr & = \frac{e}{e} \cr & = 1 \cr} $$

$$\eqalign{ & y = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right).....3n}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} \cr & \ln \,y = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\ln \left( {1 + \frac{1}{n}} \right)\left( {1 + \frac{2}{n}} \right)...\left( {1 + \frac{{2n}}{n}} \right) \cr & \ln \,y = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\ln \left( {1 + \frac{1}{n}} \right) + \ln \left( {1 + \frac{2}{n}} \right) + ..... + \ln \left( {1 + \frac{{2n}}{n}} \right)} \right] \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\ln \left( {1 + \frac{r}{n}} \right)} \cr & = \int_0^2 {\ln } \left( {1 + x} \right)dx \cr & {\text{Let }}1 + x = t\,\,\, \Rightarrow dx = dt \cr & {\text{when }}x = 0,\,\,t = 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2,\,\,\,t = 3 \cr & \ln \,y = \int_1^3 {\ln } \,t\,d\,t = \left[ {t\,\ln \,t - t} \right]_1^3 = \ln \left( {\frac{{{3^3}}}{{{e^2}}}} \right) = \ln \left( {\frac{{27}}{{{e^2}}}} \right) \cr & \Rightarrow y = \left( {\frac{{27}}{{{e^2}}}} \right) \cr} $$

$$\eqalign{ & {z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\,\sin \,\frac{{r\alpha }}{{{n^2}}}; \cr & {z_1} = \cos \frac{\alpha }{{{n^2}}} + i\,\sin \,\frac{\alpha }{{{n^2}}}; \cr & {z_2} = \cos \frac{{2\alpha }}{{{n^2}}} + i\,\sin \,\frac{{2\alpha }}{{{n^2}}};....., \cr & {z_n} = \cos \frac{{n\alpha }}{{{n^2}}} + i\,\sin \,\frac{{n\alpha }}{{{n^2}}} \cr & {\text{consider }}\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}{z_2}{z_3}.....{z_n}} \right) \cr & = \mathop {\lim }\limits_{n \to \infty } \left[ {\cos \left\{ {\frac{\alpha }{{{n^2}}}\left( {1 + 2 + 3 + ..... + n} \right)} \right\}} \right] + i\,\sin \left\{ {\frac{\alpha }{{{n^2}}}\left( {1 + 2 + 3 + ..... + n} \right)} \right\} \cr & = \mathop {\lim }\limits_{n \to \infty } \left[ {\cos \frac{{\alpha n\left( {n + 1} \right)}}{{2{n^2}}} + i\,\sin \frac{{\alpha n\left( {n + 1} \right)}}{{2{n^2}}}\,} \right] \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\cos \left( {1 + \frac{1}{n}} \right)}}{2} + \frac{{i\,\sin \,\alpha \left( {1 + \frac{1}{n}} \right)}}{2} \cr & = \cos \frac{\alpha }{2} + i\,\sin \frac{\alpha }{2} \cr & = {e^{\frac{{i\alpha }}{2}}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin \left[ {x - 3} \right]}}{{\left[ {x - 3} \right]}}} \right] \cr & {\text{For }}x \to {0^ + },\,\left[ {x - 3} \right] = - 3 \cr & \therefore \,\frac{{\sin \left[ {x - 3} \right]}}{{\left[ {x - 3} \right]}} = \frac{{\sin \left( { - 3} \right)}}{{ - 3}} = \frac{{\sin \,3}}{3} \in \left( {0,\,1} \right) \cr & \therefore \,\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin \left[ {x - 3} \right]}}{{\left[ {x - 3} \right]}} = 0 \cr & {\text{For }}x \to {0^ - },\,\left[ {x - 3} \right] = - 4 \cr & \therefore \,\frac{{\sin \left[ {x - 3} \right]}}{{\left[ {x - 3} \right]}} = \frac{{\sin \,4}}{4}{\text{ lies in }}\left( { - 1,\,0} \right) \cr & \therefore \,\mathop {\lim }\limits_{x \to {0^ - }} \left[ {\frac{{\sin \left[ {x - 3} \right]}}{{\left[ {x - 3} \right]}}} \right] = - 1 \cr & \therefore \,{\text{Limit does not exist}}{\text{.}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x\, \to \,0} \frac{{\sin \left( {\pi \,{{\cos }^2}x} \right)}}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x\, \to \,0} \frac{{\sin \left( {\pi - \pi \,{{\sin }^2}x} \right)}}{{{x^2}}}\left[ {\sin \left( {\pi - \theta } \right) = \sin \,\theta } \right] \cr & = \mathop {\lim }\limits_{x\, \to \,0} \frac{{\sin \left( {\pi \,{{\sin }^2}x} \right)}}{{\pi \,{{\sin }^2}x}} \times \frac{{\left( {\pi \,{{\sin }^2}x} \right)}}{{{x^2}}} \cr & = \pi \cr} $$

$$\eqalign{ & {\text{Given limit}} = \mathop {\lim }\limits_{x \to \alpha } \frac{{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)}}{{{{\left( {x - \alpha } \right)}^2}}} \cr & = \mathop {\lim }\limits_{x \to \alpha } \frac{{2{{\sin }^2}\left( {a\frac{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}}{2}} \right)}}{{{{\left( {x - \alpha } \right)}^2}}} \cr & = \mathop {\lim }\limits_{x \to \alpha } \frac{2}{{{{\left( {x - \alpha } \right)}^2}}} \times \frac{{{{\sin }^2}\left( {a\frac{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}}{2}} \right)}}{{\frac{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}}}{2}}} \times \frac{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}}}{4} \cr & = \frac{{{a^2}{{\left( {\alpha - \beta } \right)}^2}}}{2} \cr} $$

$$\eqalign{ & {\text{For }}n > 1, \cr & \mathop {\lim }\limits_{x \to 0} {x^n}\sin \left( {\frac{1}{{{x^2}}}} \right) = 0 \times \left( {{\text{any value between}} - 1{\text{ and }}1} \right) = 0 \cr & {\text{For }}n < 0, \cr & \mathop {\lim }\limits_{x \to 0} {x^n}\sin \left( {\frac{1}{{{x^2}}}} \right) = \infty \times \left( {{\text{any value between}} - 1{\text{ and }}1} \right) = \infty \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left| {\sin \left( {x - 1} \right)} \right|}}{{x - 1}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {\sin \,h} \right|}}{h}{\text{ or }}\mathop {\lim }\limits_{h \to 0} \frac{{\left| {\sin \,h} \right|}}{{ - h}} \cr & \therefore {\text{ RH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{h} = 1{\text{ and LH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{{ - h}} = - 1 \cr} $$