First slide

Introduction to limits

Question

If $L = \lim_{x \to 0} \frac{1}{x^{3}} (\frac{1}{\sqrt{1 + x}} - \frac{1 + ax}{1 - bx})$ exists, then

Moderate

A

$a = \frac{1}{4}$
B

$b = \frac{3}{4}$
C

$L = - \frac{1}{32}$
D

$L = \frac{1}{32}$

Solution

$\begin{array}{l} L = \lim_{x \to 0} \frac{1}{x^{3}} (\frac{1}{\sqrt{1 + x}} - \frac{1 + ax}{1 + bx}) \\ = \lim_{x \to 0} \frac{1 + bx - (1 + ax) \sqrt{1 + x}}{x^{3}} \lim_{x \to 0} \frac{1}{\sqrt{1 + x} (1 + bx)} \\ = \lim_{x \to 0} \frac{1 + bx - (1 + ax) {(1 + x)}^{\frac{1}{2}}}{x^{3}} \\ = \lim_{x \to 0} \frac{1 + bx - (a + ax) (1 + \frac{1}{2} x + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2!} x^{2} + \frac{\frac{1}{2} (\frac{1}{2} - 1) (\frac{1}{2} - 2)}{3!} x^{3} + . . .)}{x^{3}} \\ = \lim_{x \to 0} \frac{1 + bx - (1 + ax) (1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} + . . . .)}{x^{3}} \end{array}$
$\begin{array}{l} = \lim_{x \to 0} \frac{(b - \frac{1}{2} - a) x + (- \frac{a}{2} + \frac{1}{8}) x^{2} + (\frac{a}{8} - \frac{1}{16}) x^{2}}{x^{3}} \\ Limit exists, if b - \frac{1}{2} - a = 0 and - \frac{1}{8} + \frac{a}{2} = 0 \\ ∴ a = \frac{1}{4} and b = \frac{3}{4} \\ ∴ L = - \frac{1}{32} \end{array}$

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