# lim x → 0 7 x cos x − 3 sin x 4 x + tan x - Mathematics

$\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}$

#### Solution

$\lim_{x \to 0} \left[ \frac{7x \cos x - 3 \sin x}{4x + \tan x} \right]$  It is of the form $\left( \frac{0}{0} \right)$

Dividing the numerator and the denominator by x

$\Rightarrow \lim_{x \to 0} \frac{7\cos x - 3\left( \frac{\sin x}{x} \right)}{4 + \left( \frac{\tan x}{x} \right)}$
$\Rightarrow \frac{7 \lim_{x \to 0} \left( \cos x \right) - 3 \lim_{x \to 0} \left( \frac{\sin x}{x} \right)}{4 + \lim_{x \to 0} \left( \frac{\tan x}{x} \right)}$
$\Rightarrow \frac{7 . 1 - 3 . 1}{4 + 1}$
$\Rightarrow \frac{4}{5}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.7 | Q 10 | Page 50