$$L=limx$$→$$0$$ $$sin$$⁡$$x+aex+be$$−$$x+cln$$⁡$$(1+x)x3=$$∞The value of L isEquation $$ax2$$ $$+$$ bx $$+$$ c $$=$$ $$0$$ hasThe solution set of ||$$x+c$$|−$$2a$$|

Question

$$L=limx$$→$$0$$ $$sin$$⁡$$x+aex+be$$−$$x+cln$$⁡$$(1+x)x3=$$∞The value of L isEquation $$ax2$$ $$+$$ bx $$+$$ c $$=$$ $$0$$ hasThe solution set of ||$$x+c$$|−$$2a$$|<$$4b$$ is

Infinity Learn · Accepted Answer

$$L=limx$$→$$0$$ $$sin$$⁡$$x+aex+be$$−$$x+cln$$⁡$$(1+x)x3=limx$$→$$0$$ x−$$x33$$!$$+a1+x1$$!$$+x22$$!$$+x33$$!$$+b1$$−$$x1$$!$$+x22$$!−$$x33$$!$$+cx$$−$$x22+x33x3=limx$$→$$0$$ $$(a+b)+(1+a$$−$$b+c)x+a2+b2$$−$$c2x2+$$−$$13$$!$$+a3$$!−$$b3$$!$$+c3x3x3or$$ $$a+b=0$$,$$1+a$$−$$b+c=0$$,$$a2+b2$$−$$c2=0and$$ $$L=$$−$$13$$!$$+a3$$!−$$b3$$!$$+c3Solving$$ the first three equations, we get c $$=$$ $$0$$, a $$=$$ $$-1/2$$, b $$=$$ $$1/2$$.Then, L $$=$$ $$-$$ $$1/3$$.Equation $$ax2$$ $$+$$ bx $$+$$ c $$=$$ $$0$$ reduces to $$x2$$ $$-x=0$$ or $$x=0$$, $$1$$.||$$x+c$$|−$$2a$$|<$$4b$$ reduces to ∥x|$$+1$$|<$$2$$ or     −$$2$$<|x|$$+1$$<$$2$$ or     $$0$$≤|x|<$$1$$ or     x∈[−$$1$$,$$1$$]

Column -I	Column -II
A. If $L = lim_{x \to - 1} \frac{\sqrt[3]{(7 - x)} - 2}{(x + 1)}$ , then 12L =	P. -2
B. If $L = lim_{x \to π / 4} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}$ then -L/4=	Q. 2
C. If $L = lim_{x \to 1} \frac{(2 x - 3) (\sqrt{x} - 1)}{2 x^{2} + x - 3}$ , then 20L=	R. 1
D. If $L = lim_{x \to \infty} \frac{\log x^{n} - [x]}{[x]}$ , where $n \in N$ ([x] denotes greatest integer less than or equal to x),then -2L=	S. -1

Introduction to limits

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$L = lim_{x \to 0} \frac{\sin x + {ae}^{x} + {be}^{- x} + cln (1 + x)}{x^{3}} = \infty$

The value of L is

Equation ax² + bx + c = 0 has

The solution set of $| | x + c | - 2 a | < 4 b$ is

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