Proper solution of the limit of $\sin(x)/\tan(x)$ as $x \to 0$

Question

In one of my math book, I have a problem where I need to compute $\lim_{x\to0}{\frac{\sin(x)}{\tan(x)}}$

I came up with a solution that I am not able to write formally. The reasoning is the following :

• Take x as an angle of a right-angled triangle and define by "opp" the opposite side of the triangle, "adj" the adjacent side of the triangle, and "hyp" the hypothenuse of the triangle.
• $\frac{\sin(x)}{\tan(x)} = \frac{opp}{hyp}*\frac{adj}{opp} = \cos(x)$
• $\lim_{x\to0}{\frac{\sin(x)}{\tan(x)}} = \lim_{x\to0}{\cos(x)} = 1$

Is there any way to write this formally ?

Note that I am aware that I could also compute this limit by using the Taylor expansion of $\sin(x)$ and $\cos(x)$.

Answer 1

Use the fact that $$\tan(x)=\frac{\sin(x)}{\cos(x)}$$

to get $$\lim_{x\to 0} \frac{\sin(x)\cos(x)}{\sin(x)}=\lim_{x\to 0} \cos(x)$$

Your reasoning is equivalent to the first formula.

Answer 2

HINT: we have $\frac{\sin(x)}{\tan(x)}=\frac{\sin(x)}{\sin(x)} \cdot \cos(x)=\cos(x)$ and our searched limit is $1$ for $x$ tends to zero