# Lim X → 0 5 X − 1 √ 4 + X − 2 - Mathematics

$\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}$

#### Solution

$\lim_{x \to 0} \left[ \frac{5^x - 1}{\sqrt{4 + x} - 2} \right]$

Rationalising the denominator, we get:

$= \lim_{x \to 0} \left[ \frac{\left( 5^x - 1 \right) \left( \sqrt{4 + x} + 2 \right)}{\left( \sqrt{4 + x} - 2 \right) \left( \sqrt{4 + x} + 2 \right)} \right]$
$= \lim_{x \to 0} \left[ \frac{\left( 5^x - 1 \right) \left( \sqrt{4 + x} + 2 \right)}{4 + x - 4} \right]$
$= \lim_{x \to 0} \left[ \left( \frac{5^x - 1}{x} \right) \left( \sqrt{4 + x} + 2 \right) \right] \left\{ \because \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = \log a \right\}$
$= \log 5 \times \left( \sqrt{4 + 0} + 2 \right)$
$= 4 \log 5$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.1 | Q 1 | Page 71