If limx→$$0$$ $$(4x$$−$$1)13+a+bxx=13$$ , then the value of |ab|

we have limx→$$0$$ −$$(1$$−$$4x)13+a+bxx=13$$⇒limx→$$0$$ −$$1$$−$$43x+a+bxx=13$$⇒limx→$$0$$ $$(a$$−$$1)+43+bxx=13For$$ limit to exist, a $$-$$ $$1$$ $$=$$ $$0$$ or a $$=$$ $$1$$⇒limx→$$0$$ $$43+bxx=13$$⇒$$43+b=13$$⇒$$b=$$−$$1$$

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If $lim_{x \to 0} \frac{(4 x - 1)^{\frac{1}{3}} + a + bx}{x} = \frac{1}{3}$ , then the value of $| ab |$

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Solution

we have
$\begin{array}{l} lim_{x \to 0} \frac{- (1 - 4 x)^{\frac{1}{3}} + a + bx}{x} = \frac{1}{3} \\ \Rightarrow lim_{x \to 0} \frac{- [1 - \frac{4}{3} x] + a + bx}{x} = \frac{1}{3} \\ \Rightarrow lim_{x \to 0} \frac{(a - 1) + (\frac{4}{3} + b) x}{x} = \frac{1}{3} \end{array}$
For limit to exist, a - 1 = 0 or a = 1
$\begin{array}{l} \Rightarrow lim_{x \to 0} \frac{(\frac{4}{3} + b) x}{x} = \frac{1}{3} \\ \Rightarrow \frac{4}{3} + b = \frac{1}{3} \\ \Rightarrow b = - 1 \end{array}$

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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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If limx→0 (4x−1)13+a+bxx=13 , then the value of |ab|

we have limx→0 −(1−4x)13+a+bxx=13⇒limx→0 −1−43x+a+bxx=13⇒limx→0 (a−1)+43+bxx=13For limit to exist, a - 1 = 0 or a = 1⇒limx→0 43+bxx=13⇒43+b=13⇒b=−1

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If $lim_{x \to 0} \frac{(4 x - 1)^{\frac{1}{3}} + a + bx}{x} = \frac{1}{3}$ , then the value of $| ab |$