Answer
$\lim\limits_{x \to 0} \frac{xsin(x)}{1-cos(x)} = 2$
Work Step by Step
By the squeeze theorem:
$\lim\limits_{x \to 0} \frac{xsin(x)}{1-cos(x)} = \lim\limits_{x \to 0} \frac{xsin(x)}{1-cos(x)} * \frac{1+cos(x)}{1+cos(x)} = \lim\limits_{x \to 0} \frac{xsin(x)(1+cos(x))}{1-cos^2(x)} = \lim\limits_{x \to 0} \frac{xsin(x)(1+cos(x))}{sin^2(x)} = \lim\limits_{x \to 0} \frac{x(1+cos(x))}{sin(x)} = \lim\limits_{x \to 0} \frac{x}{sin(x)} * \lim\limits_{x \to 0} (1+cos(x)) = 1 * \lim\limits_{x \to 0} (1+cos(x)) = 1* (1+1) = 2$
