Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Consider }}\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^{100}}}}{{{e^x}}} + {{\left( {\cos \frac{2}{x}} \right)}^{{x^2}}}} \right] \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{100}}}}{{{e^x}}} + \mathop {\lim }\limits_{x \to \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{100}}}}{{{e^x}}} = 0\,\,\,\left( {{\text{Using L'Hospital's rule}}} \right) \cr & {\text{and }}\mathop {\lim }\limits_{x \to \infty } {\left( {\cos \frac{2}{x}} \right)^{{x^2}}}{\text{ is of }}\left( {{1^\infty }} \right){\text{ form}} \cr & = {e^{\mathop {\lim }\limits_{x \to \infty } \,{x^2}\left( {\cos \frac{2}{x} - 1} \right)}}\,\,\,\left( {{\text{Put }}\frac{2}{x} = t\,\, \Rightarrow x = \frac{2}{t}} \right) \cr & = {e^{\mathop {\lim }\limits_{t \to 0} \,\frac{4}{{{t^2}}}\left( {\cos \,t - 1} \right)}}\, \cr & = {e^{ - \mathop {\lim }\limits_{t \to 0} \left( {\frac{{1 - \cos \,t}}{{{t^2}}}} \right).4}} \cr & = {e^{\mathop { - \lim }\limits_{t \to 0} \left( {\frac{{\sin \,t}}{{2t}}} \right)4}} \cr & = {e^{ - 2}} \cr} $$

We know that
$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } {\left( {1 + x} \right)^{\frac{1}{x}}} = e \cr & \therefore \,\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2} \cr & = \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)\left( {\frac{1}{{\frac{a}{x} + \frac{b}{{{x^2}}}}}} \right)} \right]^{2x\left( {\frac{a}{x} + \frac{b}{{{x^2}}}} \right)}} = {e^2} \cr & = {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + \frac{b}{x}} \right]}} \cr & = {e^2} \cr & \Rightarrow {e^{2a}} = {e^2} \cr & \Rightarrow a = 1{\text{ and }}b \in {\bf{R}} \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x + \sqrt x - x}}{{\sqrt {x + \sqrt x } + \sqrt x }} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{\sqrt x \left\{ {\sqrt {1 + \frac{1}{{\sqrt x }}} + 1} \right\}}} \cr & = \frac{1}{2} \cr} $$

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

$$\eqalign{ & {\text{We know that }}\mathop {\lim }\limits_{x \to \infty } {\left( {1 + x} \right)^{\frac{1}{x}}} = e \cr & {\text{We have }}\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2} \cr & \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)}^{\left( {\frac{1}{{\frac{a}{x} + \frac{b}{{{x^2}}}}}} \right)}}} \right]^{2x\left( {\frac{a}{x} + \frac{b}{{{x^2}}}} \right)}} = {e^2} \cr & \Rightarrow {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + \frac{b}{x}} \right]}} = {e^2} \cr & \Rightarrow {e^{2a}} = {e^2} \cr & \Rightarrow a = 1{\text{ and}}\,b\, \in \,R \cr} $$

\[\begin{array}{l} f\left( x \right) = \sqrt {{x^2} - 10x + 25} \\ = \sqrt {{{\left( {x - 5} \right)}^2}} \\ = \left| {x - 5} \right|\\ = \left\{ \begin{array}{l} x - 5,\,\,\,\,\,x \ge 5\\ 5 - x,\,\,\,\,\,x < 5 \end{array} \right. \end{array}\]
Clearly, $$f\left( x \right)$$  is differentiable at all points on the interval $$\left[ {0,\,7} \right]$$  except at $$x = 5.$$
$$\therefore $$  The derivative of $$f\left( x \right)$$  on the interval $$\left[ {0,\,7} \right]$$  does not exist.

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 4} \frac{{4 - f\left( x \right)}}{{4 - x}}.\frac{{2 + \sqrt x }}{{2 + \sqrt {f\left( x \right)} }} \cr & = \frac{{2 + 2}}{{2 + 2}}.\mathop {\lim }\limits_{x \to 4} \frac{{4 - f\left( x \right)}}{{4 - x}} \cr & = \mathop {\lim }\limits_{x \to 4} \frac{{ - f'\left( x \right)}}{{ - 1}} \cr & = f'\left( 4 \right) \cr & = 1 \cr} $$

$$\eqalign{ & L = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}} - {{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}} \cr & \,\,\,\,\, = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( 3 \right)}^{\frac{{{x^n}}}{{{e^x}}}}}\left( {{{\left( {\frac{2}{3}} \right)}^{\frac{{{x^n}}}{{{e^x}}}}} - 1} \right)}}{{{x^n}}} \cr & {\text{Now, }}\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{e^x}}} = 0 \cr & \left( {{\text{Applying L'Hospital rule }}n{\text{ times}}} \right) \cr & {\text{Hence,}} \cr & L = \mathop {\lim }\limits_{x \to \infty } {\left( 3 \right)^{\frac{{{x^n}}}{{{e^x}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{{\left( {\frac{2}{3}} \right)}^{\frac{{{x^n}}}{{{e^x}}}}} - 1} \right)}}{{\frac{{{x^n}}}{{{e^x}}}}}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{e^x}}} \cr & \,\,\,\,\, = 1 \times \log \left( {\frac{2}{3}} \right) \times 0 \cr & \,\,\,\,\, = 0 \cr} $$

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

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{2 - \root 3 \of {8 + h} }}{{2h.\root 3 \of {8 + h} }} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{8 - \left( {8 + h} \right)}}{{2h.\root 3 \of {8 + h} \left\{ {{8^{\frac{2}{3}}} + {8^{\frac{1}{3}}}.{{\left( {8 + h} \right)}^{\frac{1}{3}}} + {{\left( {8 + h} \right)}^{\frac{2}{3}}}} \right\}}} \cr & = - \frac{1}{{48}} \cr} $$