Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \mathop {\lim }\limits_{n\, \to \,\infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + ..... + \frac{n}{{1 - {n^2}}}} \right\} \cr & = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{{1 + 2 + 3 + ..... + n}}{{1 - {n^2}}} \cr & = \mathop {\lim }\limits_{n\, \to \,\infty } \,\frac{{\sum n }}{{1 - {n^2}}} \cr & = \mathop {\lim }\limits_{n\, \to \,\infty } \,\frac{{n\left( {n + 1} \right)}}{{2\left( {1 - {n^2}} \right)}} \cr & = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{{1 + \frac{1}{n}}}{{2\left[ {\frac{1}{{{n^2}}} - 1} \right]}} = - \frac{1}{2} \cr} $$

Multiply and divide by $$x$$ in the given expression, we get
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos \,2x} \right)}}{{{x^2}}}.\frac{{\left( {3 + \cos \,x} \right)}}{1}.\frac{x}{{\tan \,4x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2\,{{\sin }^2}\,x}}{{{x^2}}}.\frac{{3 + \cos \,x}}{1}.\frac{x}{{\tan \,4x}} \cr & = 2\,\mathop {\lim }\limits_{x \to 0} \frac{{\,{{\sin }^2}\,x}}{{{x^2}}}.\mathop {\lim }\limits_{x \to 0} \,3 + \cos \,x.\mathop {\lim }\limits_{x \to 0} \,\frac{x}{{\tan \,4x}} \cr & = 2.4.\frac{1}{4}\mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{\tan \,4x}} \cr & = 2.4.\frac{1}{4} \cr & = 2 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sin \,2x + 4x}}{{2x + \sin \,4x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin \,2x}}{x} + 4}}{{2 + \frac{{\sin \,4x}}{x}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\frac{{\sin \,2x}}{{2x}}} \right) + 4}}{{2 + 4\left( {\frac{{\sin \,4x}}{{4x}}} \right)}} \cr & {\text{Applying limit, we get }}\frac{{2 + 4}}{{2 + 4}} = 1 \cr} $$

$$\eqalign{ & {\text{Given, }}\mathop {\lim }\limits_{x \to 0} \left( {{{\left( {\sin } \right)}^{\frac{1}{x}}} + {{\left( {1 + x} \right)}^{\sin \,x}}} \right) \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {\sin \,x} \right)^{\frac{1}{x}}} + \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\sin \,x}} \cr & = {\left( {\sin \,0} \right)^{\frac{1}{0}}} + \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\sin \,x}} \cr & = {\left( 0 \right)^\infty } + \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\sin \,x}} \cr & = 0 + {e^{\mathop {\lim }\limits_{x \to 0} \left[ {\log {{\left( {1 + x} \right)}^{\sin \,x}}} \right]}}\,\,\,\,\,\,\,\left( {\because \,{e^{\log \,x}} = a} \right) \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \left[ {\sin \,x\,\log \left( {1 + x} \right)} \right]}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \sin \,x \times \mathop {\lim }\limits_{x \to 0} \log \left( {1 + x} \right)}} \cr & = {e^{\sin \left( 0 \right) \times \log \left( {1 + 0} \right)}} \cr & = {e^{0 \times \log \left( 1 \right)}} \cr & = {e^{0 \times 0}} \cr & = {e^0} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Since,}}\,\,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {\frac{1}{x}} \right] + \left[ {\frac{2}{x}} \right] + ..... + \left[ {\frac{{15}}{x}} \right]} \right) \cr & = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {\frac{{1 + 2 + 3 + ..... + 15}}{x}} \right) - \left( {\left\{ {\frac{1}{x}} \right\} + \left\{ {\frac{2}{x}} \right\} + ..... + \left\{ {\frac{{15}}{x}} \right\}} \right) \cr & \because \,0 \leqslant \left\{ {\frac{r}{x}} \right\} < 1\,\,\,\,\,\,\,\, \Rightarrow \,0 \leqslant x\left\{ {\frac{r}{x}} \right\} < x \cr & \therefore \,\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\frac{{1 + 2 + 3 + ..... + 15}}{x}} \right) \cr & = \frac{{15 \times 16}}{2} \cr & = 120 \cr} $$

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

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to a} \frac{{\frac{d}{{dx}}\left\{ {x\int_a^x {f\left( x \right)dx} } \right\}}}{{\frac{d}{{dx}}\left( {x - a} \right)}} \cr & = \mathop {\lim }\limits_{x \to a} \frac{{\int_a^x {f\left( x \right)dx + xf\left( x \right)} }}{1} \cr & = af\left( a \right) \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\left[ {{1^{\frac{1}{{{{\cos }^2}x}}}} + {2^{\frac{1}{{{{\cos }^2}x}}}} + ..... + {n^{\frac{1}{{{{\cos }^2}x}}}}} \right]^{{{\cos }^2}x}} \cr & = \mathop {\lim }\limits_{t \to \infty } {\left( {{1^t} + {2^t} + ..... + {n^t}} \right)^{\frac{1}{t}}}\,\,\left[ {{\text{On putting }}\frac{1}{{{{\cos }^2}x}} = t \geqslant 1} \right] \cr & = \mathop {\lim }\limits_{t \to \infty } {\left( {{n^t}} \right)^{\frac{1}{t}}}{\left[ {{{\left( {\frac{1}{n}} \right)}^t} + {{\left( {\frac{2}{n}} \right)}^t} + ..... + {{\left( {\frac{n}{n}} \right)}^t}} \right]^{\frac{1}{t}}} \cr & = n\,\mathop {\lim }\limits_{t \to \infty } {\left[ {{{\left( {\frac{1}{n}} \right)}^t} + {{\left( {\frac{2}{n}} \right)}^t} + ..... + {{\left( {\frac{n}{n}} \right)}^t}} \right]^{\frac{1}{t}}} \cr & = n{\left( {0 + 0 + ..... + 1} \right)^0} \cr & = n \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{{{4^{\frac{1}{n}}} - 1}}{{{3^{\frac{1}{n}}} - 1}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{{4^{\frac{1}{n}}} - 1}}{{\frac{1}{n}}}}}{{\frac{{{3^{\frac{1}{n}}} - 1}}{{\frac{1}{n}}}}} \cr & {\text{Since }}n{\text{ tends to }}\infty \frac{1}{n}\,{\text{tends to 0}} \cr & {\text{So, applying }}\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = {\log _e}a \cr & = \frac{{{{\log }_e}4}}{{{{\log }_e}3}} = {\log _e}4 \cr} $$

$$\eqalign{ & f\left( x \right) = x,\,\,g\left( x \right) = \frac{1}{x} \cr & \mathop {\lim }\limits_{x \to 0} f\left( x \right) = 0,\,\,\mathop {\lim }\limits_{x \to 0} g\left( x \right) = {\text{does not exist}} \cr & {\text{But }}\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \mathop {\lim }\limits_{x \to 0} \left[ {{x^2}} \right] = 0 \cr} $$
Hence, none of these is only true option.