Why can we calculate trigonometric limits by substitution?

Question

I have always wondered why something like these $\displaystyle \lim_{x\to 0} \sin x=0$ and $\displaystyle \lim_{x\to 0} \cos x=1$ are true. Why should this happen? Can anyone explain this to me with rigor? Thanks.

Answer 1

The fact that $\sin x\to 0$ is easily seen geometrically if you know the statement: "The shortest curve from the point $(x,y)$ to the $x$ axis is of length $|y|$."

This can be used to show that $|\sin x|\leq |x|$, since the arc of the circle from $(\cos x,\sin x)$ to the point $(1,0)$ if length $|x|$ is a path to the $x$-axis, and that has to be at larger than $|\sin x|$. This in turn shows that $\sin x\to 0$ as $x\to 0$.

For $\cos x$, use $\cos^2 x + \sin^2 x = 1$ and that $\cos x$ is positive for $x$ small. Since $\sin x\to 0$, we have $\cos ^2 x\to 1$. Since $\cos x$ is positive for small $x$, this means that $\cos x\to 1$.

Answer 2

For the first limit, it follows from inequality $|\sin x|\leq| x|$ as you can deduce by a geometric argument on the unit circle. The second limit will follows from the first since $\cos x=\sqrt {1-\sin^2 x} $ in a neightborhood of 0.

Answer 3

Let me try to answer this question myself with the hints given above.

Theorem: $\sin x$ is continuous on $\mathbb{R}$

Proof: For a circle of radius 1, the length of the circumference is $2\pi$ and so the length of the arc subtended by the angle $x$ is $\displaystyle \frac{2\pi}{2\pi}x=x$. $|x|\ge |\sin x|$ as evident from the diagram.

For $\sin x$ to be continuous on $\mathbb{R}$,we need to show that for $a\in \mathbb{R}$, given $\epsilon>0 \exists , \delta>0:|x-a|<\delta\implies |\sin x-\sin a|<\epsilon$. $$|\sin x-\sin a|=|2\sin\frac{x-a}{2}\cos\frac{x+a}{2}|\leq |2\sin\frac{x-a}{2}|\leq|2\frac{x-a}{2}| $$ uing the above result. Take $\delta=\epsilon$.

QED.

Using this,$$\lim_{x\to 0}\sin x=0$$ Is my answer correct?