Limits MCQ Questions & Answers in Calculus | Maths

By using the rule, $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \,x}}{x} = 1$$
We get, required limit $$ = \frac{{10}}{9}.\frac{9}{{10}}.\frac{8}{7}.\frac{7}{8}.\frac{6}{5}.\frac{5}{6}.\frac{4}{3}.\frac{3}{4}.\frac{1}{2} = \frac{1}{2}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right).\left( {1 - \sin \,x} \right)}}{{{{\left( {\pi - 2x} \right)}^3}}} \cr & {\text{Let }}x = \frac{\pi }{2} + y;\,\,\,\,\,y \to 0 \cr & = \mathop {\lim }\limits_{y \to 0} \frac{{\tan \left( { - \frac{y}{2}} \right).\left( {1 - \cos \,y} \right)}}{{{{\left( { - 2y} \right)}^3}}} \cr & = \mathop {\lim }\limits_{y \to 0} \frac{{ - \tan \,\frac{y}{2}2\,si{n^2}\,\frac{y}{2}}}{{\left( { - 8} \right).\frac{{{y^3}}}{8}.8}} \cr & = \mathop {\lim }\limits_{y \to 0} \frac{1}{{32}}\frac{{\tan \,\frac{y}{2}}}{{\left( {\frac{y}{2}} \right)}}.{\left[ {\frac{{\frac{{\sin \,y}}{2}}}{{\frac{y}{2}}}} \right]^2} \cr & = \frac{1}{{32}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{{{5^{n + 1}} + {3^n} - {2^{2n}}}}{{{5^n} + {2^n} + {3^{2n + 3}}}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{{{5.5}^n} + {3^n} - {4^n}}}{{{5^n} + {2^n} + {{27.9}^n}}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{5.\frac{{{5^n}}}{{{9^n}}} + \frac{{{3^n}}}{{{9^n}}} - \frac{{{4^n}}}{{{9^n}}}}}{{\frac{{{5^n}}}{{{9^n}}} + \frac{{{2^n}}}{{{9^n}}} + 27}} \cr & = \frac{{0 + 0 - 0}}{{0 + 0 + 27}} \cr & = 0 \cr} $$

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

$$f\left( x \right)$$   is a positive increasing function.
$$\eqalign{ & \therefore 0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right) \cr & \Rightarrow 0 < 1 < \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} < \frac{{f\left( {3x} \right)}}{{f\left( x \right)}} \cr & \Rightarrow \mathop {\lim }\limits_{x \to \infty } \,1 \leqslant \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} \leqslant \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {3x} \right)}}{{f\left( x \right)}} \cr} $$
By Sandwich Theorem.
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} = 1$$

$$\eqalign{ & {\text{As }}x \to 0 - \left( {{\text{i}}{\text{.e}}{\text{., approaches 0 from the left}}} \right),\,\left[ x \right] = - 1 \cr & \therefore \,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{1 + \sin \left( { - 1} \right)}}{{ - 1}} = - 1 + \sin \,1 \cr & {\text{Whereas, if }}x \to {0^ + }{\text{ we get }}\left[ x \right] = 0 \cr & \therefore \,f\left( x \right) = 0\, \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 0 \cr & {\text{Thus, }}\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ does not exist}}{\text{.}} \cr} $$

Consider
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \,{{\cos }^2}\,x} \right)}}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]}}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \,\sin \frac{{\left( {\pi - \pi {{\sin }^2}x} \right)}}{{{x^2}}}\,\,\,\,\,\,\,\,\,\left[ {\because \sin \,\left( {\pi - \theta } \right) = \sin \,\theta } \right] \cr & = \mathop {\lim }\limits_{x \to 0} \,\sin \frac{{\left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \times \frac{{\pi {{\sin }^2}x}}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \,1 \times \pi {\left( {\frac{{\sin x}}{x}} \right)^2} \cr & = \pi \cr} $$

KEY CONCEPT:
$$\eqalign{ & \frac{d}{{dx}}\left[ {\int_{g\,\left( x \right)}^{h\,\left( x \right)} {f\left( t \right)dt} } \right] \cr & = f\left( {h\left( x \right)} \right).h'\left( x \right) - f\left( {g\left( x \right).g'\left( x \right)} \right) \cr & = \frac{{f\left( 2 \right) \times 2 \times 2 \times 1}}{{2 \times \frac{\pi }{4}}} = \frac{8}{\pi }f\left( 2 \right) \cr & {\text{Let }}L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\int\limits_2^{{{\sec }^2}\,x} {f\left( t \right)dt} }}{{{x^2} - \frac{{{\pi ^2}}}{{16}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\frac{0}{0}{\text{ form}}} \right] \cr} $$
On applying L'Hospital's rule, we get
$$\eqalign{ & L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\frac{d}{{dx}}\left[ {\int\limits_2^{{{\sec }^2}\,x} {f\left( t \right)dt} } \right]}}{{\frac{d}{{dx}}\left( {{x^2} - \frac{{{\pi ^2}}}{{16}}} \right)}}\,\,\, \cr & L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{f\left( {{{\sec }^2}\,x} \right).2\,{{\sec }^2}\,x\,\tan \,x}}{{2x}}\,\,\, \cr} $$

$$\eqalign{ & {x_{n + 1}} = \sqrt {2 + {x_n}} \cr & \Rightarrow \lim {x_{n + 1}} = \sqrt {2 + \lim \,{x_n}} \cr & \Rightarrow t = \sqrt {2 + t} \,\,\,\,\,\left[ {\because \,\,\lim \,{x_{n + 1}} = \lim \,{x_n}} \right] \cr & \Rightarrow {t^2} - t - 2 = 0 \cr & \Rightarrow \left( {t - 2} \right)\left( {t + 1} \right) = 0 \cr & \Rightarrow t = 2 \cr} $$

$$\eqalign{ & {\text{Let }}\left[ a \right] = n,\,{\text{then }}\mathop {\lim }\limits_{x \to {n^ - }} \frac{{{e^{\left\{ x \right\}}} - \left\{ x \right\} - 1}}{{{{\left\{ x \right\}}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{\left\{ {n - h} \right\}}} - \left\{ {n - h} \right\} - 1}}{{{{\left\{ {n - h} \right\}}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{1 - h}} - \left( {1 - h} \right) - 1}}{{{{\left( {1 - h} \right)}^2}}} \cr & = e - 2 \cr & {\text{and }}\mathop {\lim }\limits_{x \to {n^ + }} \frac{{{e^{\left\{ x \right\}}} - \left\{ x \right\} - 1}}{{{{\left\{ x \right\}}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{\left\{ {n + h} \right\}}} - \left\{ {n + h} \right\} - 1}}{{{{\left\{ {n + h} \right\}}^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - h - 1}}{{{h^2}}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 + h + \frac{{{h^2}}}{{2!}} + \frac{{{h^3}}}{{3!}}..... - h - 1}}{{{h^2}}} = \frac{1}{2} \cr & \therefore {\text{ Limits does not exist}}{\text{.}} \cr} $$