Proof of Trig Limits — Question 10

Question 10 - Solution

We begin by rewriting the expression to isolate a known limit.

Factor out $x$ from the numerator: $$\frac{\tan x - x}{x} = \frac{\tan x}{x}-1.$$

Now rewrite $\tan x$ in terms of sine and cosine: $$\frac{\tan x}{x} = \frac{\sin x}{x\cos x} = \left(\frac{\sin x}{x}\right)\left(\frac{1}{\cos x}\right).$$

Take limits of each factor as $x\to 0$.

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

Since cosine is continuous at $0$, $$\lim_{x\to 0}\cos x=\cos 0=1, \quad \text{so} \quad \lim_{x\to 0}\frac{1}{\cos x}=1.$$

Thus, $$\lim_{x\to 0}\frac{\tan x}{x}=1.$$

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

$$\boxed{0}$$