Infinite Limits — Question 2

Question 2 - Solution

We evaluate $$\lim_{x \to 0^+} \frac{\tan x}{x}.$$

Method 1: Local Approximation

As $x \to 0$, the Taylor expansion of $\tan x$ is $$\tan x = x + \frac{x^3}{3} + \cdots.$$ Thus, $$\frac{\tan x}{x} = \frac{x + \frac{x^3}{3} + \cdots}{x} = 1 + \frac{x^2}{3} + \cdots.$$

As $x \to 0^+$, the higher-order terms approach 0, so $$\lim_{x \to 0^+} \frac{\tan x}{x} = 1.$$

Method 2: Rewrite Using Sine and Cosine

Recall that $$\tan x = \frac{\sin x}{\cos x}.$$ Then, $$\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}.$$

Now evaluate each factor as $x \to 0^+$: $$\lim_{x \to 0^+} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0^+} \cos x = 1.$$

Therefore, $$\lim_{x \to 0^+} \frac{\tan x}{x} = 1 \cdot \frac{1}{1} = 1.$$

Conclusion

$$\boxed{\lim_{x \to 0^+} \frac{\tan x}{x} = 1}$$