L'Hospital rule simplification

Question

How does the text simplify from

$$\frac{1}{3} \lim_{x \to 0} \frac{\tan{x}}{x} $$

$$ = \frac{1}{3} \lim_{x \to 0} \frac{\sec^2{x}}{1}$$

How does that work?

First step (I think?):

$$ = \frac{1}{3} \lim_{x \to 0} \frac{\sin{x}}{\cos{x} \cdot {x}}$$

Answer 1

Note that since $\cos x\to 1$ we have

$$\frac{1}{3} \lim_{x \to 0} \frac{\sin{x}}{\cos{x} \cdot {x}}= \frac{1}{3} \lim_{x \to 0} \frac{1}{\cos{x}}\cdot \lim_{x \to 0} \frac{\sin{x}}{x}=\frac13\cdot 1\cdot 1$$

indeed

$$\lim_{x \to 0} \frac{\sin{x}}{x}=1$$

which is a well known standard limit or to which apply l’Hopital

$$\lim_{x \to 0} \frac{\sin{x}}{x}=\lim_{x \to 0} \frac{\cos{x}}{1}=1$$

Answer 2

The derivative of $\tan x$ is $\sec^{2}x$, and the derivative of $x$ is $1$. When we apply L'Hopital rule, we do derivatives in the both numerator and denominator.