Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{We have,}} \cr & f\left( x \right) = {\left( {{{\sin }^{m - n}}x} \right)^{m + n}}.{\left( {{{\sin }^{n - p}}x} \right)^{n + p}}.{\left( {{{\sin }^{p - m}}x} \right)^{p + m}} \cr & = {\sin ^{{m^2} - {n^2}}}x.{\sin ^{{n^2} - {p^2}}}x.{\sin ^{{p^2} - {m^2}}}x \cr & = {\left( {\sin \,x} \right)^{{m^2} - {n^2} + {n^2} - {p^2} + {p^2} - {m^2}}} \cr & = {\left( {\sin \,x} \right)^0} \cr & = 1 \cr & \therefore \,\,f'\left( x \right) = 0 \cr} $$

$$\eqalign{ & \because \,\left[ {\frac{1}{x}} \right] = {\text{ Integer}} \cr & \therefore \,{\left( { - 1} \right)^{\left[ {\frac{1}{x}} \right]}} = 1{\text{ or }} - 1 \cr & \mathop {\lim }\limits_{x \to 0} x{\left( { - 1} \right)^{\left[ {\frac{1}{x}} \right]}} \cr & = \mathop {\lim }\limits_{h \to 0} \left( h \right)\left( {1{\text{ or }} - 1} \right) = 0 \cr & = \mathop {\lim }\limits_{h \to 0} \left( { - h} \right)\left( {1{\text{ or }} - 1} \right) = 0 \cr} $$

\[\begin{array}{l} g\left\{ {f\left( x \right)} \right\} = \left\{ \begin{array}{l} {\left\{ {f\left( x \right)} \right\}^2} + 1,\,f\left( x \right) \ne 2\\ 3,\,f\left( x \right) = 2 \end{array} \right.\\ \therefore g\left\{ {f\left( x \right)} \right\} = \left\{ \begin{array}{l} {\sin ^2}x + 1,\,x \ne n\pi \\ 3,\,x = n\pi \end{array} \right. \end{array}\]
$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{x \to 0 + 0} g\left\{ {f\left( x \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} g\left\{ {f\left( {0 + h} \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin }^2}\left( {0 + h} \right) + 1} \right\} \cr & = 1 \cr} $$
$$\eqalign{ & {\text{LH limit}} = \mathop {\lim }\limits_{x \to 0 - 0} g\left\{ {f\left( x \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} g\left\{ {f\left( {0 - h} \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin }^2}\left( {0 - h} \right) + 1} \right\} \cr & = 1 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos \,2x} }}{{\sqrt 2 x}}\,\, \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}x} \right)} }}{{\sqrt 2 x}}; \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2\,{{\sin }^2}x} }}{{\sqrt 2 x}}\,\,\, \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\left| {\sin \,x} \right|}}{x} \cr} $$
The limit of above does not exist as
$${\text{LHS}} = - 1 \ne \,\,{\text{RHL}} = 1$$

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \left[ {\root 3 \of {{{\left( {n + 1} \right)}^2}} - \root 3 \of {{{\left( {n - 1} \right)}^2}} } \right] \cr & = \mathop {\lim }\limits_{n \to \infty } {n^{\frac{2}{3}}}\left[ {{{\left( {1 + \frac{1}{n}} \right)}^{\frac{2}{3}}} - {{\left( {1 - \frac{1}{n}} \right)}^{\frac{2}{3}}}} \right] \cr & = \mathop {\lim }\limits_{n \to \infty } {n^{\frac{2}{3}}}\left[ {\left( {1 + \frac{2}{3}.\frac{1}{n} + \frac{{\frac{2}{3}\left( {\frac{2}{2} - 1} \right)}}{{2!}}\frac{1}{{{n^2}}}.....} \right) - \left( {1 - \frac{2}{3}.\frac{1}{n} + \frac{{\frac{2}{3}\left( {\frac{2}{3} - 1} \right)}}{{2!}}\frac{1}{{{n^2}}}.....} \right)} \right] \cr & = \mathop {\lim }\limits_{n \to \infty } {n^{\frac{2}{3}}}\left[ {\frac{4}{3}.\frac{1}{n} + \frac{8}{{81}}.\frac{1}{{{n^3}}} + .....} \right] \cr & = \left[ {\frac{4}{3}.\frac{1}{{{n^{\frac{1}{3}}}}} + \frac{8}{{81}}.\frac{1}{{{n^{\frac{7}{3}}}}} + .....} \right] \cr & = 0 \cr} $$

$$\mathop {\lim }\limits_{n \to \infty } f\left( {\frac{n}{{\sqrt {9{n^2} + 1} }}} \right) = f\left\{ {\mathop {\lim }\limits_{n \to \infty } \frac{n}{{\sqrt {9{n^2} + 1} }}} \right\} = f\left( {\frac{1}{3}} \right) = 1$$

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]{\text{is equal to}} \cr & \mathop {\lim }\limits_{n \to \infty } \frac{r}{{{n^2}}}{\sec ^2}\frac{{{r^2}}}{{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}.\frac{r}{n}{\sec ^2}\frac{{{r^2}}}{{{n^2}}} \cr & \Rightarrow {\text{Given limit is equal to value of integral}} \cr & \int_0^1 {x\,{{\sec }^2}{x^2}dx = \frac{1}{2} \times } \int_0^1 {2x\,{{\sec }^2}{x^2}dx} \cr & {\text{Put }}{x^2} = t, \cr & {\text{The integral becomes}} = \frac{1}{2}\int_0^1 {{{\sec }^2}t\,dt = \frac{1}{2}\left( {\tan \,t} \right)_0^1 = \frac{1}{2}\tan \,1} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0 + } \left[ {\frac{{\sin \left( {\operatorname{sgn} \,x} \right)}}{{\operatorname{sgn} \left( x \right)}}} \right] = \mathop {\lim }\limits_{x \to 0 + } \left[ {\frac{{\sin \,1}}{1}} \right] = 0 \cr & = \mathop {\lim }\limits_{x \to {0^ - }} \left[ {\frac{{\sin \left( {\operatorname{sgn} \,x} \right)}}{{\operatorname{sgn} \left( x \right)}}} \right] = \mathop {\lim }\limits_{x \to {0^ - }} \left[ {\frac{{\sin \left( { - 1} \right)}}{{ - 1}}} \right] = \mathop {\lim }\limits_{x \to {0^ - }} \left[ {\sin \,1} \right] \cr & {\text{Hence, the given limit is 0}}{\text{. }} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{{{\log }_e}\left[ x \right] \times \frac{x}{x}}}{x} \cr & {\text{as }}x \to \infty \cr & \frac{{\left[ x \right]}}{x} = 1 \cr & {\text{We know that,}} \cr & \left( {\frac{{x - 1}}{x} \leqslant \frac{{\left[ x \right]}}{x} \leqslant \frac{x}{x} \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{\left[ x \right]}}{x} = 1} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\log }_e}\left[ x \right]}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \,x}}{x}\,\,\,\,\,\,\,\,\,\,\frac{\infty }{\infty }{\text{ from}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} \times \frac{1}{1} = \frac{1}{\infty } = 0 \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2.\frac{{{x^2}}}{{1 + 3{x^2}}}} \right)^{{{\frac{1}{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + 2.\frac{1}{{3 + \frac{1}{{{x^2}}}}}} \right)^{\left( {3 + \frac{1}{{{x^2}}}} \right).\frac{{\frac{1}{{{x^2}}}}}{{3 + \frac{1}{{{x^2}}}}}}} \cr & = {\left\{ {\mathop {\lim }\limits_{y \to \infty } {{\left( {1 + \frac{2}{y}} \right)}^y}} \right\}^{\mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{1}{{{x^2}}}}}{{3 + \frac{1}{{{x^2}}}}}}} \cr & = {\left( {{e^2}} \right)^{\mathop {\lim }\limits_{x \to 0} \,\frac{1}{{3{x^2} + 1}}}} \cr & = {e^2} \cr} $$