Question

lim x tends to 0 3sinx-4sin^3/x

Answer 1

it's the formula for sin 3theta/x
I mean sin3x/x
and we know that sinx/x=1
so ans is 3

Answer 2

$\lim_{x \to 0} \dfrac{3\sin x-4\sin^3 x}{x}=3$

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{3\sin x-4\sin^3 x}{x}$

To find, $\lim_{x \to 0} \dfrac{3\sin x-4\sin^3 x}{x}=?$

∴ $\lim_{x \to 0} \dfrac{3\sin x-4\sin^3 x}{x}$ ......(1)

($\dfrac{0}{0}$ form)

We know that,

$\sin 3A=3\sin A-4\sin^3 A$

Equation (1) can be written as,

$\lim_{x \to 0} \dfrac{\sin 3x}{x}$

Multiplying numerator and denominator by 3, we get

=$\lim_{x \to 0} \dfrac{3\sin 3x}{3x}$

$=3\lim_{x \to 0} \dfrac{\sin 3x}{3x}$

= 3 × 1

Using limit formula,

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

= 3

∴ $\lim_{x \to 0} \dfrac{3\sin x-4\sin^3 x}{x}=3$

Hence, $\lim_{x \to 0} \dfrac{3\sin x-4\sin^3 x}{x}=3$