Question

limit as x tends to 0 : (4sin x - 9x cosx ) / (3x^2-5 tan x)​

Answer 1

Answer:

1

Step-by-step explanation:

use la hospitals rule

Answer 2

The value of $\lim_{x \to 0} \dfrac{4\sin x - 9x\cos x}{3x^2- \tan x}$ = 1

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{4\sin x - 9x\cos x}{3x^2- \tan x}$

To find, the value of $\lim_{x \to 0} \dfrac{4\sin x - 9x\cos x}{3x^2- \tan x}$ = ?

∴ $\lim_{x \to 0} \dfrac{4\sin x - 9x\cos x}{3x^2- \tan x}$ ( $\dfrac{0}{0}$ form)

Dividing numerator and denominator by x, we get

$=\lim_{x \to 0} \dfrac{\dfrac{4\sin x - 9x\cos x}{x} }{\dfrac{3x^2- \tan x}{x}}$

$=\lim_{x \to 0} \dfrac{4\dfrac{\sin x }{x}- 9\cos x }{3x-5\dfrac{\tan x}{x}}$

We know that,

$\lim_{x \to 0} \dfrac{\sin x}{x}=1$ and

$\lim_{x \to 0} \dfrac{\tan x}{x}=1$

$=\lim_{x \to 0} \dfrac{4(1)- 9\cos x }{3x-5(1)}$

Put x = 0, we get

$=\dfrac{4- 9\cos 0 }{3(0)-5}$

$=\dfrac{4- 9(1) }{3(0)-5}$

$=\dfrac{-5 }{-5}$

= 1

Thus, the value of $\lim_{x \to 0} \dfrac{4\sin x - 9x\cos x}{3x^2- \tan x}$ = 1