Limit of Special Trigonometric

Problem 239: Proof that

(1) $\begin{equation*} \lim_{x \to 0} \frac{\cos x - 1}{x} = 0. \end{equation*}$

Solution: We will start by multiplying the numerator and denominator by the conjugate $\cos x + 1$ . That is,

(2) $\begin{align*} \lim_{x \to 0} \frac{\cos x - 1}{x} & = \lim_{x \to 0} \frac{\cos x - 1}{x} \cdot \frac{\cos x + 1}{\cos x + 1} \\ & = \lim_{x \to 0} \frac{\cos^2 x - \cos x + \cos x - 1}{x \cos x + x} \\ & = \lim_{x \to 0} \frac{\cos^2 x - 1}{x\cos x + x} \\ & = \lim_{x \to 0} \frac{1-\cos^2 x}{-x\cos x - x}.\end{align*}$

Let’s recall that $\sin^2 x + \cos^2 x = 1$ so that $\sin^2 x = 1-\cos^2$ . Thus

(3) $\begin{align*} \lim_{x \to 0} \frac{1-\cos^2 x}{-x\cos x - x} &= \lim_{x \to 0} \frac{\sin^2 x}{-x\cos x - x} \\ & = \lim_{x \to 0}\frac{\sin x \cdot \sin x}{-x(\cos x + 1)}\\ & = \lim_{x \to 0} \left(-\frac{\sin x}{x} \right) \cdot \left(\frac{\sin x}{1+\cos x} \right) \\ & = \underbrace{\left( \lim_{x \to 0} -\frac{\sin x}{x} \right)}_{= -1} \cdot \underbrace{\left(\lim_{x \to 0} \frac{\sin x}{1 + \cos x}\right)}_{= 0} \\ & = -1 \cdot 0 \\& = 0. \end{align*}$

This completes the proof.

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