Question

evaluate lim x-> 0
(4 sin x - 9x cosx) / (3x² - 5 tanx)

Answer 1

value of given limit is 1

evaluate Lim (x → 0) (4sinx - 9x cosx)/(3x² - 5tanx)

putting x = 0,

(4 × sin0 - 9 × 0 × cos0)/(3.0² - 5tan0) = 0/0 so it is the form of limit.

we know, lim(x→0) = sinx/x = 1 and lim(x →0) = tanx/x = 1 let's apply these.

Lim(x→0) {4(sinx/x) × x - 9xcosx}/{3x² - 5(tanx/x)x}

= lim(x→0) (4x - 9x cosx)/(3x² - 5x)

= lim(x →0) (4 - 9cosx)/(3x - 5)

putting x = 0,

= (4 - 9cos0)/(3 × 0 - 5)

= (4 - 9)/(-5)

= -5/-5

= 1

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Answer 2

$\huge\bold\purple{Answer:-}$

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