Question

Find the limit when X approaches zero 2xsinx/1-cosx

Answer 1

Answer:

4

Step-by-step explanation:

$Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}$

$=Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}\times \frac{1+\cos x}{1+\cos x}$

$=Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{1^2 -\cos^2 x}$

$=Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{1 -\cos^2 x}$

$=Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{sin^2 x}$

$=Lim_{x \to 0}\frac{2x(1+\cos x) }{sin x}$

$=Lim_{x \to 0} 2(1+\cos x) \times \frac{1}{Lim_{x \to 0}\frac{sin x}{x}}$

$=2(1+\cos 0) \times 1$

$= 2(1+1)$

$= 2(2)$

$\therefore Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}= 4$