Question

evaluate Lt x_0
$\sin3x - \sinx \div \sin4x - \sin2x$
​

Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sin(3x)-\sin x}{\sin(4x)-\sin(2x)}\quad\quad\dots(1)$

We have,

• $\sin A-\sin B=2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$

The numerator is,

$\longrightarrow\sin(3x)-\sin x=2\cos\left(\dfrac{3x+x}{2}\right)\sin\left(\dfrac{3x-x}{2}\right)$

$\longrightarrow\sin(3x)-\sin x=2\cos\left(2x\right)\sin x$

And the denominator is,

$\longrightarrow\sin(4x)-\sin(2x)=2\cos\left(\dfrac{4x+2x}{2}\right)\sin\left(\dfrac{4x-2x}{2}\right)$

$\longrightarrow\sin(4x)-\sin(2x)=2\cos\left(3x\right)\sin x$

Then (1) becomes,

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{2\cos(2x)\sin x}{2\cos(3x)\sin x}$

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\cos(2x)}{\cos(3x)}$

Now putting $x=0,$

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\cos(2\cdot0)}{\cos(3\cdot0)}$

$\displaystyle\longrightarrow\underline{\underline{L=1}}$

Hence 1 is the answer.