Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\bf{L}}{\bf{.H}}{\bf{.L}}{\bf{.}} = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {p + 1} \right)h}}{{ - h}} + \frac{{\sin \left( { - h} \right)}}{{ - h}} \cr & = p + 1 + 1 \cr & = p + 2 \cr & {\bf{R}}{\bf{.H}}{\bf{.L}} = \mathop {\lim }\limits_{x \to {\sigma ^ + }} f\left( x \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 + h} - 1}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {\sqrt {1 + h} + 1} \right)}} \cr & = \frac{1}{2} \cr & {\text{and }}f\left( 0 \right) = q\,\,\,\,\, \Rightarrow p = - \frac{3}{2},\,\,q = \frac{1}{2} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} = \mathop {\lim }\limits_{x \to {0^ + }} {\left( x \right)^0} = 1,\, \cr & \mathop {\lim }\limits_{x \to {0^ - }} {\left( { - x} \right)^0} = \mathop {\lim }\limits_{x \to {0^ - }} \,1 = 1 \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\sec }^2}x}}{{\tan \,x + x\,{{\sec }^2}x}}{\text{ (using L'Hospital rule)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{{\sec }^2}x.\tan \,x}}{{2{{\sec }^2}x + x.2{{\sec }^2}x.\tan \,x}} \cr & = \frac{0}{2} = 0 \cr} $$

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

$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \cr & {\text{ = }}\mathop {\lim }\limits_{x \to 0 - } f\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left[ {\cos \,h} \right]}}{{1 + \left[ {\cos \,h} \right]}}{\text{ = }}\frac{{\sin \left( 0 \right)}}{{1 + 0}} = 0 \cr & \left( {\therefore h > 0 \Rightarrow \cos \,h < 1} \right) \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} \cr & {\text{ = }}\mathop {\lim }\limits_{x \to 0 + } f\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left[ {\cos \,h} \right]}}{{1 + \left[ {\cos \,h} \right]}}{\text{ = }}\frac{{\sin \left( 0 \right)}}{{1 + 0}} = 0 \cr} $$

$$\eqalign{ & {\text{We have}} = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} \cr & = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\frac{r}{{n\sqrt {1 + {{\left( {\frac{r}{n}} \right)}^2}} }}} \cr & = \int_0^2 {\frac{x}{{\sqrt {1 + {x^2}} }}dx} \,\,\,\left[ {\because \mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{r}\sum\limits_{r\, = \,0}^{{a_n}} {f\left( {\frac{r}{n}} \right) = \int\limits_0^a {f\left( x \right)dx} } } \right] \cr & = \left[ {\sqrt {1 + {x^2}} } \right]_0^2 = \sqrt 5 - 1 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 2}}} \right)^x} \cr & = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{{4x + 1}}{{{x^2} + x + 2}}} \right)^x} \cr & = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + \frac{{4x + 1}}{{{x^2} + x + 2}}} \right)}^{\frac{{{x^2} + x + 2}}{{4x + 1}}}}} \right]^{\frac{{\left( {4x + 1} \right)x}}{{{x^2} + x + 2}}}} \cr & = \mathop {\lim }\limits_{{e^{x \to \infty }}} \frac{{4{x^2} + x}}{{{x^2} + x + 2}} \cr & = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{{4 + \frac{1}{x}}}{{1 + \frac{1}{x} + \frac{2}{{{x^2}}}}}}} \cr & = {e^4} \cr} $$

$$\eqalign{ & {\text{For }}x \in R, \cr & \mathop {\lim }\limits_{x\, \to \,\infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x\, \to \,\infty } {\left\{ {{{\left[ {1 - \frac{5}{{x + 2}}} \right]}^{\frac{{ - \,\left( {x + 2} \right)}}{5}}}} \right\}^{\frac{{ - \,5x}}{{x + 2}}}} \cr & = {e^{\mathop {\lim }\limits_{x \to \,\infty } - \,\frac{5}{{1\, + \,\frac{2}{x}}}}} = {e^{ - \,5}} \cr} $$

$$\eqalign{ & {x^2} + 2x + 3 = {\left( {x + 1} \right)^2} + 2 \geqslant 2 \cr & {\text{So, }}a = 2;\,\,b = \mathop {\lim }\limits_{\theta \to 0} \frac{{2\,{{\sin }^2}\frac{\theta }{2}}}{{{\theta ^2}}} = \frac{1}{2} \cr & \therefore \sum\limits_{r = 0}^n {{a^r}{b^{n - r}} = {b^n} + a{b^{n - 1}} + {a^2}{b^{n - 2}} + ..... + {a^n}} \cr & = \frac{{{b^n}\left\{ {1 - {{\left( {\frac{a}{b}} \right)}^{n + 1}}} \right\}}}{{1 - \frac{a}{b}}} \cr & = \frac{{{{\left( {\frac{1}{2}} \right)}^n}\left\{ {1 - {4^{n + 1}}} \right\}}}{{1 - 4}} \cr & = \frac{{{4^{n + 1}} - 1}}{{3 \times {2^n}}} \cr} $$

From the question, $$f\left( a \right) = 0$$   and $$f\left( x \right)$$  is differentiable at $$x=a$$
$$\therefore {\text{Limit}} = \mathop {\lim }\limits_{x \to a} \frac{{\frac{1}{{1 + 6f\left( x \right)}}.6f'\left( x \right)}}{{3f'\left( x \right)}} = 2.\frac{1}{{1 + 6f\left( a \right)}} = 2$$