Limits MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 1} \sum\limits_{r = 1}^n {\frac{{{x^r} - {1^r}}}{{x - 1}}} \cr & = \sum\limits_{r = 1}^n {r \cdot {1^{r - 1}}} \cr & = 1 + 2 + 3 + ..... + n \cr & = \frac{{n\left( {n + 1} \right)}}{2} \cr} $$

$${\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \frac{{1 + {{\left( {\frac{b}{a}} \right)}^n}}}{{1 - {{\left( {\frac{b}{a}} \right)}^n}}} = 1$$      because $$0 < \frac{b}{a} < 1$$   implies $${\left( {\frac{b}{a}} \right)^n} \to 0$$   as $$n \to \infty $$

$$\eqalign{ & {\text{We have,}} \cr & y = \left( {1 + {x^{\frac{1}{4}}}} \right)\left( {1 + {x^{\frac{1}{2}}}} \right)\left( {1 - {x^{\frac{1}{4}}}} \right) \cr & = \left( {1 - {x^{\frac{1}{2}}}} \right)\left( {1 + {x^{\frac{1}{2}}}} \right) \cr & = 1 - x \cr & \therefore \,\frac{{dy}}{{dx}} = - 1 \cr} $$

$${\text{Limit}} = {\left( {2 - 1} \right)^{\log \,1}} = {1^0} = 1$$

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

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {\left[ {1 + x\,\ell n{{\left( {1 + b} \right)}^2}} \right]^{\frac{1}{x}}} = 2b\,{\sin ^2}\theta \cr & \Rightarrow {e^{\mathop {\lim }\limits_{x \to 0} \,\frac{1}{x}\,\ell n\left[ {1 + x\,\ell n{{\left( {1 + b} \right)}^2}} \right]}} = 2b\,{\sin ^2}\theta \cr & \Rightarrow {e^{\mathop {\lim }\limits_{x \to 0} \,\frac{{\ell n\left[ {1 + x\,\ell n\,\left( {1 + {b^2}} \right)} \right]}}{{x\,\ell n\,\left( {1 + {b^2}} \right)}} \times \ell n\,\left( {1 + {b^2}} \right)}} = 2b\,{\sin ^2}\theta \cr & \Rightarrow {e^{\ell n\,\left( {1 + {b^2}} \right)}} = 2b\,{\sin ^2}\theta \cr & \Rightarrow 1 + {b^2} = 2b\,{\sin ^2}\theta \cr & \Rightarrow 2\,{\sin ^2}\theta = b + \frac{1}{b} \cr} $$
We know that $$2\,{\sin ^2}\theta \leqslant 2$$     and $$b + \frac{1}{b} \geqslant 2$$   for $$b>0$$
$$\eqalign{ & \therefore 2\,{\sin ^2}\theta = b + \frac{1}{b} = 2\,\,\,\,\,\,\, \Rightarrow {\sin ^2}\theta = 1 \cr & {\text{As }}\theta \in \left( { - \pi ,\,\,\pi } \right], \cr & \therefore \theta = \pm \frac{\pi }{2} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 2 - 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 2 - 0} \left( {{x^2} - 1} \right) = {2^2} - 1 = 3 \cr & \mathop {\lim }\limits_{x \to 2 + 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 2 + 0} \left( {2x + 3} \right) = 2 \times 2 + 3 = 7 \cr} $$
$$\therefore $$  Required quadratic equation is $${x^2} - 10x + 21 = 0$$

$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ {2 - \left( {2 + h} \right)} \right] + \left[ {\left( {2 + h} \right) - 2} \right] - \left( {2 + h} \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ { - h} \right] + \left[ h \right] - 2 - h} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ { - 1 + 0 - 2 - h} \right\} \cr & = - 3 \cr & {\text{LH limit}} = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ {2 - \left( {2 - h} \right)} \right] + \left[ {\left( {2 - h} \right) - 2} \right] - \left( {2 - h} \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ h \right] + \left[ { - h} \right] - 2 + h} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {0 - 1 - 2 + h} \right\} \cr & = - 3 \cr} $$

$$\eqalign{ & \because \,\,\frac{\pi }{4} < 1,\,\,\,\therefore \,\left[ {\frac{\pi }{4}} \right] = 0 \cr & \therefore \,\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {\frac{x}{2}} \right]}}{{\ln \left( {\sin \,x} \right)}} = 0 \cr} $$

$$\eqalign{ & {\text{As }}0 \leqslant x - \left[ x \right] < 1\,\forall \,x\, \in \,R,\,0 \leqslant f\left( x \right) < 1 \cr & \therefore \,\mathop {\lim }\limits_{n \to \infty } {\left\{ {f\left( x \right)} \right\}^{2n}} = 0 \cr & {\text{Thus, for}}\,x\, \in \,R, \cr & g\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left\{ {f\left( x \right)} \right\}}^{2n}} - 1}}{{{{\left\{ {f\left( x \right)} \right\}}^{2n}} + 1}} = \frac{{0 - 1}}{{0 + 1}} = - 1{\text{ }} \cr} $$