Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {\left( {\cos \,x} \right)^{\frac{1}{{\sin \,x}}}}{\text{ is }}{1^\infty }{\text{form}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \left( {\cos \,x - 1} \right)\frac{1}{{\sin \,x}}}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{ - 2\,{{\sin }^2}\frac{x}{2}}}{{2\,\sin \frac{x}{2}\cos \frac{x}{2}}}}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \left( { - \tan \frac{x}{2}} \right)}} \cr & = {e^o} \cr & = 1 \cr} $$

$${\text{We have, }}\mathop {\lim }\limits_{x \to 2} \frac{{x\,f\left( 2 \right) - 2\,f\left( x \right)}}{{x - 2}}\,\,\left( {\frac{0}{0}} \right)$$
By applying 'L'Hospital rule, we get
$$\eqalign{ & = \mathop {\lim }\limits_{x \to 2} f\left( 2 \right) - 2f'\left( x \right) \cr & = f\left( 2 \right) - 2f'\left( 2 \right) \cr & = 4 - 2 \times 4 \cr & = - 4 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \frac{\pi}{2}} \frac{{\cot \,x\left( {1 - \sin \,x} \right)}}{{ - 8{{\left( {x - \frac{\pi }{2}} \right)}^3}}} \cr & = \mathop {\lim }\limits_{x \to \frac{\pi}{2}} \frac{{\cot \,x\left( {1 - \sin \,x} \right)}}{{8{{\left( {\frac{\pi }{2} - x} \right)}^3}}} \cr & {\text{put }}\frac{\pi }{2} - x = t\,\,\,\, \Rightarrow {\text{as }}x \to \frac{\pi }{2}\,\,\,\, \Rightarrow t \to 0 \cr & = \mathop {\lim }\limits_{t \to 0} \frac{{\cot \left( {\frac{\pi }{2} - t} \right)\left( {1 - \sin \left( {\frac{\pi }{2} - t} \right)} \right)}}{{8{t^3}}} \cr & = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t\left( {1 - \cos \,t} \right)}}{{8{t^3}}} \cr & = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t}}{{8t}}.\frac{{1 - \cos \,t}}{{{t^2}}} \cr & = \frac{1}{8}.1.\frac{1}{2} \cr & = \frac{1}{{16}} \cr} $$

$$\eqalign{ & {\text{Let }}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sin \,nx} \right)\left[ {\left( {a - n} \right)nx - \tan \,x} \right]}}{{{x^2}}} = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\left( {nx - \frac{{{n^3}{x^3}}}{{3!}}} \right)\left[ {n\left( {a - n} \right)x - \left\{ {x + \frac{{{x^3}}}{3} + .....} \right\}} \right]}}{{{x^2}}} = 0 \cr & \left( {{\text{By using expansion of sin }}x{\text{ and tan }}x} \right) \cr & \Rightarrow {n^2}\left( {a - n} \right) - n = 0 \cr & \Rightarrow an - {n^2} - 1 = 0 \cr & \Rightarrow a = \frac{{{n^2} + 1}}{n} \cr & \Rightarrow a = n + \frac{1}{n} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{{1 + 2 + 3 + ..... + n}}{{{1^2} + {2^2} + {3^2} + ..... + {n^2}}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}} \cr & \therefore \,\,\mathop {\lim }\limits_{n \to \infty } \frac{3}{{2n + 1}} = 0 \cr & {\bf{Note:}}{\text{ }}1 + 2 + 3 + ..... + n = \frac{{n\left( {n + 1} \right)}}{2} \cr & {1^2} + {2^2} + {3^2} + ..... + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,\sqrt h }}{{\sqrt {\sin \,h} }} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sin \,\sqrt h }}{{\sqrt h }}} \right)\sqrt h \left( {\frac{1}{{\sqrt {\frac{{\sin \,h}}{h}} }}} \right)\frac{1}{{\sqrt h }} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{We have,}} \cr & \frac{d}{{dx}}\left[ {\frac{{\left( {1 + {x^2} + {x^4}} \right)\left( {1 - {x^2} + {x^4}} \right)}}{{\left( {1 + {x^2} + {x^4}} \right)}}} \right] = a{x^3} + bx \cr & \Rightarrow \frac{d}{{dx}}\left( {1 - {x^2} + {x^4}} \right) = a{x^3} + bx \cr & \Rightarrow - 2x + 4{x^3} = a{x^3} + bx \cr & \Rightarrow a = 4{\text{ and }}{\mkern 1mu} b = - 2{\text{ }} \cr} $$

$$\mathop {\lim }\limits_{x \to 0} f\left( {\frac{{1 - \cos \,3x}}{{{x^2}}}} \right) = f\left\{ {\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \,3x}}{{{x^2}}}} \right\} = f\left( {\frac{9}{2}} \right) = \frac{2}{9}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 - x} \right)\left( {1 - {x^2}} \right).....\left( {1 - {x^{2n}}} \right)}}{{{{\left\{ {\left( {1 - x} \right)\left( {1 - {x^2}} \right).....\left( {1 - {x^n}} \right)} \right\}}^2}}} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\frac{{1 - x}}{{1 - x}}} \right)\left( {\frac{{1 - {x^2}}}{{1 - x}}} \right).....\left( {\frac{{1 - {x^{2n}}}}{{1 - x}}} \right)}}{{{{\left( {\left( {\frac{{1 - x}}{{1 - x}}} \right)\left( {\frac{{1 - {x^2}}}{{1 - x}}} \right).....\left( {\frac{{1 - {x^n}}}{{1 - x}}} \right)} \right)}^2}}} \cr & = \frac{{1 \times 2 \times 3.....\left( {2n} \right)}}{{{{\left( {1 \times 2 \times 3.....n} \right)}^2}}} \cr & = \frac{{\left( {2n} \right)!}}{{n!n!}} \cr & = {}^{2n}{C_n} \cr} $$

$$\eqalign{ & {\text{Consider }}\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^2}}}{{3x - 2}} - \frac{x}{3}} \right] \cr & = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{3{x^2} - x\left( {3x - 2} \right)}}{{3\left( {3x - 2} \right)}}} \right] \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{2x}}{{3\left( {3x - 2} \right)}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{2}{3}\frac{1}{{\left( {3 - \frac{2}{x}} \right)}} \cr & = \frac{2}{9} \cr} $$