# Other ways to evaluate $\lim_{\theta\to 0} \frac{\sin2\theta}{2\theta}$?

What steps should be taken to find the limit: $$\lim_{\theta\to 0} \frac{\sin2\theta}{2\theta}$$?

I went about evaluating the limit using the fundamental rules of limits. I noticed that $\lim_{\theta\to 0}$ $\sin\theta\over\theta$ $=1$ and that the $\lim_{x\to a}$ $cx=ca$

Therefore, this meant that the limit of the function evaluated will go as followed, $\sin2\theta\over2 \theta$ $\to$ $2\times1\over2$ $\to$ $1$

I was wondering if I evaluated the function correctly, or did I get to this by luck? Is there another way to evaluate this equation?

• No step at all: the limit of $\dfrac{\sin mx}{nx}$ as $x$ tends to $0$ ($m/n$) is a limit every well-bred young people should know by heart. – Bernard Jan 2 '17 at 23:19
• @Bernard: Why? I can't think of an occasion in which knowledge of this limit has been useful to me. Indeed, though I see that the result is true and I have seen it before (I think in an exercise in Spivak), I can't even recall an occasion in which I've seen it used. – Will R Jan 3 '17 at 1:16
• Do you prefer to have to rewrite it: $\;\dfrac{\sin mx}{mx}\times\dfrac{mx}{nx}$ ad nauseam? – Bernard Jan 3 '17 at 1:26

Another way is to note that $$\lim_{\theta\to 0}\frac{\sin2\theta}{2\theta} =\lim_{\theta\to 0}\frac{2\sin\theta\cos\theta}{2\theta} =\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\lim_{\theta\to 0}{\cos\theta} =(1)(1)=1$$

If $\theta \to 0$, then $2\theta\to 0$ hence by setting $x:=2\theta$ we have that $$\lim\limits_{\theta \to 0}\frac{\sin (2\theta)}{2\theta} = \lim\limits_{x\to0}\frac{\sin(x)}{x} = 1.$$

(As noted in the comments, this substitution is justified, since it being a continuous substitution is sufficient for everything to be alright.)

• You're calculating the limit as $\theta\to0$ of the composite function $f(g(\theta))$ where $f(x):=\frac{\sin x}{x}$ and $g(\theta):=2\theta$ by setting $x:=g(\theta)$ and calculating the limit as $x\to0$ of $f(x)$. In general such a change of variable will not give the same limit. Under appropriate conditions, it will indeed give the same limit, but I don't see where you have noticed such conditions in your answer. – Guest Jan 2 '17 at 23:10
• @Guest: ??? This is a continuous change of variable! – Bernard Jan 2 '17 at 23:17
• @Bernard I'm referring to this result. Let $f(1):=3$ and $f(y):=2$ if $y\neq1$ and $g(x):=1$. Reading this answer, it would seem that since $g(x)\to1$ as $x\to0$, then the limit as $x\to0$ of $f(g(x))$ would equal the limit as $y\to1$ of $f(y)$, which is false. Is my objection not justified? – Guest Jan 2 '17 at 23:23
• Yes. This is the (rare) case when $g(x)$ takes the forbidden value an infinite number of times. This does not happen here, and the domain $\dfrac{\sin x}x$ can be extended by continuity at $0$. – Bernard Jan 2 '17 at 23:31
• @Bernard I think the answer should add such details (even though I've never seen anyone quote that result or check any of the two conditions it gives to justify such a change of variables... and I don't know why). – Guest Jan 2 '17 at 23:36

Well, since very fortunately L'Hopital rule is not prohibited here, why don't we use it:

$$\lim_{x\to 0 }\frac{\sin (2x)}{2x}=\lim_{x\to 0}\frac{2\cos 2x}{2}=1.$$

• This seems a bit murky to me, since the limit is simply $\frac 12 f'(0)$ where $f(x) = \sin 2x$. Stuff like this is why I'm uncomfortable using L'hopital. – MathematicsStudent1122 Jan 3 '17 at 1:13
• @MathematicsStudent1122: Well, as long as it is not circular, then there is no harm to use it. – Jack Jan 3 '17 at 1:16