Proof of Trig Limits — Question 6

Question 6 - Solution

We begin by factoring the expression.

Rewrite the fraction as $$\frac{\sin x - x}{x} = \frac{\sin x}{x}-1.$$

Now take the limit as $x\to 0$.

Using the fundamental trigonometric limit, $$\lim_{x\to 0}\frac{\sin x}{x}=1.$$

Therefore, $$\lim_{x\to 0}\left(\frac{\sin x}{x}-1\right) = 1-1 = 0.$$

Hence, $$\boxed{\lim_{x\to 0}\frac{\sin x - x}{x}=0}.$$