First slide

Introduction to limits

Question

$If lim_{x \to 0} \frac{(4 x - 1)^{\frac{1}{3}} + a + b x}{x} exists and is equal to \frac{1}{3}, then a b =$

Difficult

A

1
B

$\frac{1}{2}$
C

$- 1$
D

$- \frac{1}{2}$

Solution

$\begin{array}{l} lim_{x \to 0} \frac{(4 x - 1)^{\frac{1}{3}} + a + b x}{x} = \frac{1}{3} \\ \Rightarrow lim_{x \to 0} \frac{- [1 - \frac{4}{3} x] + a + b x}{x} = \frac{1}{3} [∵ {(1 - 4 x)}^{\frac{1}{3}} i s i n t h e f o r m o f {(1 - x)}^{\frac{p}{q}}] \\ \Rightarrow lim_{x \to 0} \frac{(a - 1) + (\frac{4}{3} + b) x}{x} = \frac{1}{3} \\ For limit to exist, a = 1 \\ \Rightarrow lim_{x \to 0} \frac{(\frac{4}{3} + b) x}{x} = \frac{1}{3} \\ \Rightarrow b = - 1 \end{array}$

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