Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\tan ^2}x\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} - \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right) \cr & = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\tan ^2}x\frac{{\left( {2\,{{\sin }^2}x + 3\sin \,x + 4 - \,{{\sin }^2}x - 6\sin \,x - 2} \right)}}{{\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} }} \cr & = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\tan }^2}x\left( {{{\sin }^2}x - 3\sin \,x + 2} \right)}}{{\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} }} \cr & = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\sin }^2}x\left( {\sin \,x - 1} \right)\left( {\sin \,x - 2} \right)}}{{\left( {1 - {{\sin }^2}x} \right)\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right)}} \cr & = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{ - {{\sin }^2}x\left( {\sin \,x - 2} \right)}}{{\left( {1 - \sin x} \right)\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right)}} \cr & = \frac{1}{{2\left( {\sqrt 9 + \sqrt 9 } \right)}} \cr & = \frac{1}{{12}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2 - \sqrt {1 + \cos \,x} }} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{9^x}{{.3}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2 - \sqrt 2 \,\cos \frac{x}{2}}} \cr & = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{9^x} - 1}}{x}} \right).\left( {\frac{{{3^x} - 1}}{x}} \right).\frac{1}{{\sqrt 2 }}.{x^2}.\frac{1}{{2\,{{\sin }^2}\frac{x}{4}}} \cr & = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{9^x} - 1}}{x}} \right).\left( {\frac{{{3^x} - 1}}{x}} \right).\frac{1}{{\sqrt 2 }}\left( {\frac{{\frac{{{x^2}}}{{16}}}}{{{{\sin }^2}\frac{x}{4}}}} \right)8 \cr & = \frac{8}{{\sqrt 2 }}\left( {\log \,9} \right)\left( {\log \,3} \right) \cr & = 8\sqrt 2 {\left( {\log \,3} \right)^2} \cr} $$

$$\eqalign{ & {\text{Given that, }}x > 0\,{\text{and }}g\left( x \right){\text{ is bounded function}}{\text{.}} \cr & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right){e^{nx}} + g\left( x \right)}}{{{e^{nx}} + 1}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right)}}{{1 + \left( {\frac{1}{{{e^{nx}}}}} \right)}} + \frac{{g\left( x \right)}}{{{e^{nx}} + 1}} \cr & = \frac{{f\left( x \right)}}{{1 + 0}} + \frac{{{\text{finite}}}}{\infty } \cr & = f\left( x \right) \cr} $$

$$\eqalign{ & y = \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right) \cr & \frac{{dy}}{{dx}} = \left( { - \frac{1}{{{x^2}}}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right) + \left( {1 + \frac{1}{x}} \right)\left( { - \frac{2}{{{x^2}}}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right) + ... + \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( { - \frac{n}{{{x^2}}}} \right) \cr & \because \,\,{\left. {\frac{{dy}}{{dx}}} \right|_{x = - 1}} = \left( { - 1} \right)\left( { - 1} \right)\left( { - 2} \right)\left( { - 3} \right).....\left( {1 - n} \right) \cr & = {\left( { - 1} \right)^n}\left( 1 \right)\left( 2 \right)\left( 3 \right).....\left( {n - 1} \right) \cr & = {\left( { - 1} \right)^n}\left( {n - 1} \right)! \cr} $$

We have
$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}}; \cr & \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r\, = \,1}^n {\frac{{{r^p}}}{{{n^p}.n}} = \int\limits_0^1 {{x^p}dx = \left[ {\frac{{{x^{p + 1}}}}{{p + 1}}} \right]_0^1 = \frac{1}{{p + 1}}} } \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \,x\,\sin \left( {\frac{1}{x}} \right) + \sin \left( {\frac{1}{{{x^2}}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\sin \left( {\frac{1}{x}} \right)}}{{\left( {\frac{1}{x}} \right)}} + \mathop {\lim }\limits_{x \to \infty } \,\sin \left( {\frac{1}{{{x^2}}}} \right) \cr & = 1 + 0 \cr & = 1 \cr} $$

$$\eqalign{ & \ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{2x}}\ln \left( {1 + {{\tan }^2}\,\sqrt x } \right) \cr & \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x}\ln \left( {\sec \sqrt x } \right) \cr} $$
Applying L hospital's rule :
$$\eqalign{ & = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sec \sqrt x .\tan \sqrt x }}{{\sec \sqrt x .2\sqrt x }} \cr & = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\tan \sqrt x }}{{2\sqrt x }} \cr & = \frac{1}{2} \cr} $$

Since, $$f\left( x \right)$$  is a polynomial function satisfying
$$\eqalign{ & f\left( x \right).f\left( {\frac{1}{x}} \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right), \cr & \therefore \,f\left( x \right) = {x^n} + 1{\text{ or }}f\left( x \right) = - {x^n} + 1 \cr & {\text{If }}f\left( x \right) = - {x^n} + 1,{\text{ then }}f\left( 4 \right) = - {4^n} + 1 \ne 65 \cr & {\text{So, }}f\left( x \right) = {x^n} + 1{\text{ Since, }}f\left( 4 \right) = 65 \cr & \therefore \,{4^n} + 1 = 65\,\, \Rightarrow \,n = 3 \cr & \therefore \,f\left( x \right) = {x^3} + 1\,\, \Rightarrow f'\left( x \right) = 3{x^2} \cr & \therefore \,f'\left( {{l_1}} \right) = 3l_1^2,\,f'\left( {{l_2}} \right) = 3l_2^2,\,f'\left( {{l_3}} \right) = 3l_3^2 \cr & {\text{Since, }}{l_1},\,{l_2},\,{l_3}{\text{ are in GP}}{\text{.}} \cr & \therefore \,f'\left( {{l_1}} \right),\,f'\left( {{l_2}} \right),\,f'\left( {{l_3}} \right){\text{ are also in GP}}{\text{.}} \cr} $$

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

$$\eqalign{ & \mathop {\lim }\limits_{h \to 0} \left[ {{\text{cose}}{{\text{c}}^3}x \cdot \cot \,x - 2\,{{\cot }^3}x \cdot {\text{cosec}}\,x + \frac{{{{\cot }^4}x}}{{\sec \,x}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos \,x}}{{{{\sin }^4}x}} - \frac{{2\,{{\cos }^3}x}}{{{{\sin }^4}x}} + \frac{{{{\cos }^5}x}}{{{{\sin }^4}x}}} \right) \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\cos \,x{{\left( {1 - {{\cos }^2}x} \right)}^2}}}{{{{\sin }^4}x}} \cr & = \mathop {\lim }\limits_{x \to 0} \cos \,x \cr & = 1 \cr} $$