Limits MCQ Questions & Answers in Calculus | Maths

We first find the derivatives of $$f\left( x \right)$$  at $$x = - 1$$   and at $$x = 0.$$
We have,
$$\eqalign{ & f'\left( { - 1} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( { - 1 + h} \right) - f\left( { - 1} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2{{\left( { - 1 + h} \right)}^2} + 3\left( { - 1 + h} \right) - 5} \right] - \left[ {2{{\left( { - 1} \right)}^2} + 3\left( { - 1} \right) - 5} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} - h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {2h - 1} \right) \cr & = 2\left( 0 \right) - 1 \cr & = - 1 \cr & {\text{and }}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2{{\left( {0 + h} \right)}^2} + 3\left( {0 + h} \right) - 5} \right] - \left[ {2{{\left( 0 \right)}^2} + 3\left( 0 \right) - 5} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} + 3h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {2h + 3} \right) \cr & = 2\left( 0 \right) + 3 \cr & = 3 \cr & {\text{Clearly, }}f'\left( 0 \right) = - 3f'\left( { - 1} \right) \cr} $$

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

$$\eqalign{ & f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } n\left( {{x^{\frac{1}{n}}} - 1} \right) \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{{x^{\frac{1}{n}}} - 1}}{{\frac{1}{n}}} \cr & = \mathop {\lim }\limits_{m \to 0} \frac{{{x^m} - 1}}{m} = {\text{ln}}\,x\,\,\left( {{\text{where }}\frac{1}{n}\,{\text{is replaced by }}m} \right) \cr & {\text{or }}f\left( {xy} \right) = \ln \left( {xy} \right) = \ln \,x + \ln \,y = f\left( x \right) + f\left( y \right) \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \alpha } \frac{{1 - \cos \left( {a{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}} \cr & = \mathop {\lim }\limits_{x \to \alpha } \frac{{2\,{{\sin }^2}\left( {\frac{{a{x^2} + bx + c}}{2}} \right)}}{{{{\left( {x - \alpha } \right)}^2}}} \cr & = \mathop {\lim }\limits_{x \to \alpha } 2{\left[ {\frac{{\sin \frac{{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}}{2}}}{{\frac{{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}}{2}}}} \right]^2} \times \frac{{{a^2}{{\left( {x - \beta } \right)}^2}}}{4} \cr & = \frac{{{a^2}}}{2}{\left( {\alpha - \beta } \right)^2}\,\,\,\left[ {{\text{using }}a{x^2} + bx + c = a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \right] \cr} $$

$${\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{\sin \frac{a}{n}}}{{\frac{a}{n}}}.\frac{a}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{a}{n}} \right)^n} = {e^a}$$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{x\,{{\log }_e}3}} - 1}}{x}.\left( {\sqrt {x + 1} + 1} \right) \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\left\{ {1 + x\,{{\log }_e}3 + {{\left( {\frac{{x\,{{\log }_e}3}}{{2!}}} \right)}^2} + .....} \right\} - 1}}{x}.\left( {\sqrt {x + 1} + 1} \right) \cr & = \left( {{{\log }_e}3} \right) \times 2 \cr & = {\log _e}9 \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{h \to 0} \frac{{\int_1^{1 + h} {\left| {t - 1} \right|dt} }}{{\sin \left( {1 + h - 1} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{d}{{dh}}\int_1^{1 + h} {\left| {t - 1} \right|dt} }}{{\frac{d}{{dh}}\left( {\sin \,h} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {1 + h - 1} \right|.\frac{{d\left( {1 + h} \right)}}{{dh}}}}{{\cos \,h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left| h \right|}}{{\cos \,h}} \cr & = 0 \cr} $$