Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left[ {0 + h} \right]}}{{\left[ {0 + h} \right]}} = \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \,\theta }}{\theta } = 1 \cr & {\text{LH limit}} = \mathop {\lim }\limits_{k \to 0} \frac{{\sin \left[ {0 - h} \right]}}{{\left[ {0 - h} \right]}} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( { - 1} \right)}}{{ - 1}} = \sin \,1 \cr & \therefore \,{\text{RH limit}} \ne {\text{LH limit}} \cr} $$

$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 + h} \right)\sin \left\{ {1 + h - \left[ {1 + h} \right]} \right\}}}{{1 + h - 1}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 + h} \right)\sin \left( {1 + h - 1} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {1 + h} \right)\frac{{\sin \,h}}{h} \cr & = 1 \cr & {\text{LH limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 - h} \right)\sin \left\{ {1 - h - \left[ {1 - h} \right]} \right\}}}{{1 - h - 1}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 - h} \right)\sin \left( {1 - h} \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - h}}{{ - h}}.\sin \left( {1 - h} \right) \cr & = - \infty \times \sin \,1 \cr & \therefore \,{\text{RH limit}}\,\, \ne \,\,{\text{LH limit}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ......+\frac{{{x^2}}}{2} + x + 1 \cr & \Rightarrow f'\left( x \right) = \frac{{100{x^{99}}}}{{100}} + \frac{{99{x^{98}}}}{{99}} + ...... + \frac{{2x}}{2} + 1 + 0 \cr & \left[ {\because \,f\left( x \right) = {x^n} \Rightarrow f'\left( x \right) = n{x^{n - 1}}} \right] \cr & \Rightarrow f'\left( x \right) = {x^{99}} + {x^{98}} + ..... + x + 1......({\text{i}}) \cr & {\text{Putting, }}x = 1{\text{ we get}} \cr & f'\left( 1 \right) = \underbrace {{{\left( 1 \right)}^{99}} + {{\left( 1 \right)}^{98}} + ...... + 1 + 1}_{100\,{\text{ times}}} = \underbrace {1 + 1 + 1 + ...... + 1 + 1}_{100\,{\text{ times}}} \cr & \Rightarrow f'\left( 1 \right) = 100......({\text{ii}}) \cr & {\text{Again, putting}}\,x = 0,{\text{ we get}} \cr & f'\left( 0 \right) = 0 + 0 + ...... + 0 + 1 \cr & \Rightarrow f'\left( 0 \right) = 1......({\text{iii}}) \cr & {\text{From equation (ii) and (iii), we get; }}f'\left( 1 \right) = 100f'\left( 0 \right) \cr & {\text{Hence,}}\,m = 100 \cr} $$

$$\eqalign{ & {\text{We have,}} \cr & y = \frac{1}{{1 + \frac{{{x^\beta }}}{{{x^\alpha }}} + \frac{{{x^\gamma }}}{{{x^\alpha }}}}} + \frac{1}{{1 + \frac{{{x^\alpha }}}{{{x^\beta }}} + \frac{{{x^\gamma }}}{{{x^\beta }}}}} + \frac{1}{{1 + \frac{{{x^\alpha }}}{{{x^\gamma }}} + \frac{{{x^\beta }}}{{{x^\gamma }}}}} \cr & = \frac{{{x^\alpha }}}{{{x^\alpha } + {x^\beta } + {x^\gamma }}} + \frac{{{x^\beta }}}{{{x^\alpha } + {x^\beta } + {x^\gamma }}} + \frac{{{x^\gamma }}}{{{x^\alpha } + {x^\beta } + {x^\gamma }}} \cr & = \frac{{{x^\alpha } + {x^\beta } + {x^\gamma }}}{{{x^\alpha } + {x^\beta } + {x^\gamma }}} \cr & = 1 \cr & \therefore \,\frac{{dy}}{{dx}} = 0 \cr} $$

$${\text{Limit}} = \sqrt {\mathop {\lim }\limits_{x \to 0} \frac{{x - \sin \,x}}{{x + {{\sin }^2}x}}} = \sqrt {\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \,x}}{{1 + \sin \,2x}}} = 0$$

$$\eqalign{ & f\left( x \right) = \sqrt {x - 1} + \sqrt {25 + \left( {x - 1} \right) - 10\sqrt {x - 1} } \cr & = \sqrt {x - 1} + \sqrt {{{\left( {5 - \sqrt {x - 1} } \right)}^2}} \cr & = \sqrt {x - 1} + \left| {5 - \sqrt {x - 1} } \right| \cr & = 5\,\,\,\left[ {\because \,\sqrt {x - 1} < 5{\text{ for }}1 < x < 26} \right] \cr & \therefore \,f'\left( x \right) = 0 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)}}{x} = K\,\,\,\left[ {{\text{by L'Hospital rule}}} \right] \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{3 + x}} - \frac{{ - 1}}{{3 - x}}}}{1} = K \cr & \therefore \frac{2}{3} = K \cr} $$

$$\eqalign{ & {\text{Given that }}f:R \to R{\text{ be such that }}f\left( 1 \right) = 3{\text{ and }}f'\left( 1 \right) = 6 \cr & {\text{Then }}\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{f\left( {1 + x} \right)}}{{f\left( 1 \right)}}} \right)^{\frac{1}{x}}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left[ {\log \,f\left( {1 + x} \right) - \log \,f\left( 1 \right)} \right]}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{f\left( {1 + x} \right)}}f'\left( {1 + x} \right)}}{1}}} \cr & = {e^{\frac{{f'\left( 1 \right)}}{{f\left( 1 \right)}}}} \cr & = {e^{\frac{6}{3}}} \cr & = {e^2} \cr} $$
[ Using L Hospital rule ]

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} }}{{\frac{d}{{dx}}\left( {x\,\sin \,x} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}{x^2}.2x}}{{\sin \,x + x\,\cos \,x}}\,\,\,\,\left[ {{\text{By L'Hospital rule}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sec }^2}{x^2}}}{{\left( {\frac{{\sin \,x}}{x} + \cos \,x} \right)}} \cr & = \frac{{2 \times 1}}{{1 + 2}} \cr & = 1 \cr} $$

We are given that
$$\mathop {\lim }\limits_{x \to 0} \frac{{\left[ {\left( {a - n} \right)nx - \tan \,x} \right]\sin \,nx}}{{{x^2}}} = 0$$
where $$n$$ is non zero real number
$$\eqalign{ & \Rightarrow \mathop {\lim }\limits_{x \to 0} n.\frac{{\sin \,nx}}{{nx}}\left[ {\left\{ {\left( {a - n} \right)n - \frac{{\tan \,x}}{x}} \right\}} \right] = 0 \cr & \Rightarrow 1.n\left[ {\left( {a - n} \right)n - 1} \right] = 0 \cr & \Rightarrow a = \frac{1}{n} + n\,\,{\text{or}}\,\,\,n + \frac{1}{n} \cr} $$