Answer:
4
Step-by-step explanation:
[tex] Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x} [/tex]
[tex] =Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}\times \frac{1+\cos x}{1+\cos x} [/tex]
[tex] =Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{1^2 -\cos^2 x} [/tex]
[tex] =Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{1 -\cos^2 x} [/tex]
[tex] =Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{sin^2 x} [/tex]
[tex] =Lim_{x \to 0}\frac{2x(1+\cos x) }{sin x} [/tex]
[tex] =Lim_{x \to 0} 2(1+\cos x) \times \frac{1}{Lim_{x \to 0}\frac{sin x}{x}} [/tex]
[tex] =2(1+\cos 0) \times 1 [/tex]
[tex] = 2(1+1) [/tex]
[tex] = 2(2) [/tex]
[tex] \therefore Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}= 4 [/tex]
