trigonometry - How lim $h->0$ $sin h/h$ is equal to $1$?

How $\lim_{h\to 0} \sin h$ when divided by $h$ gives the value $1$ ? Does it also follows with other ratios that for example. $\lim_{h\to 0} \cos x/x = 1$

Answer

One way to think about this is to recall the relationship between the trig functions and circles (in particular and wlog, the unit circle).

First, we're measuring angles by arc length subtended (i.e., in radians). The theorem comes from the fact that when the angle is very small (call it $\theta$), then $\sin\theta\approx\theta\approx\tan\theta$ (draw a diagram!). More correctly, though, what holds is that $$\sin\theta<\theta<\tan\theta$$ for $\theta<π/2$, so that we have, upon division by $\sin\theta$ and taking reciprocals, that $$\cos\theta<\sin\theta/\theta<1.$$ Now letting $\theta \to 0$, we obtain the result (a similar argument can be made from the other direction).

No, it doesn't hold for the other trig functions similarly. In particular, $$\lim_{\theta\to 0}{\cos\theta/\theta}=\infty.$$