# About $\lim_{x\rightarrow 0}\frac {\sin x}{x} = 1$ [duplicate]

I do not understand how $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ As if $$x = 0, \frac{\sin (0)}{0} = \frac {0}{0}$$ So if someone could explain this I would appreciate it! Thanks!

## marked as duplicate by Javier, egreg, Hans Lundmark, Davide Giraudo, GuyMar 29 '14 at 19:00

• Use L'Hospital's rule or the taylor series of sin(x). – Peter Mar 29 '14 at 18:13
• When you have a limit, you do not care what happens at the value $0$. You only care about what happens near $0$. – Sujaan Kunalan Mar 29 '14 at 18:13
• @JavierBadia I think the main issue here is not understanding that a limit is not simply evaluating the function at the point; less a question of why $\lim_{x\rightarrow 0}{\frac{\sin x}{x}}=1$. – Hayden Mar 29 '14 at 18:35

You can think of $\lim\limits_{x\to 0}f(x)=L$ to mean that as you approach $f(x)$ from the left of $0$ and the right of $0$, the function looks as if it is going to be that value, however it does not necessarily have to be that value. See below.

Let this curve be $g(x)$. $\lim\limits_{x\to 2}g(x)$ exists (because if I look at $g(x)$ from the left of $2$ and the right of $2$ it looks like its going to equal about $1$) but it is not equal to $g(2)$, which is actually $1.5$.

Extra Example:

Say I gave you a discontinuous function, such as $f(x)=\frac{1}{x}$, and I asked you to find $\lim\limits_{x\to 0}\frac{1}{x}$. Even without the graph, you can probably tell that this function is continuous everywhere except for $x=0$. (An easy way to see this is if you recall that division by zero is undefined and see all the ways that the denominator can possibly be $0$). For $f(x)=\frac{1}{x}$, only $x=0$ would make $\frac{1}{x}$ undefined.

First off, if $\lim_{x\to 0}\frac{1}{x}$ exists, then $\lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)$, which just means that if I approach $f(x)$ from the left of $0$ or the right of $0$, I should get the same answer.

Now $\lim\limits_{x\to 0^+}\frac{1}{x}$ can be thought of as as $x$ gets smaller and smaller, what are the values of $\frac{1}{x}$ going to approach? Well, $\frac{1}{10}=0.1<\frac{1}{1}=1<\frac{1}{0.1}=10<\frac{1}{0.01}=100<\ldots <\frac{1}{0.000001}=1000000$.

As the $x$ values get smaller and smaller from the right of $0$, $f(x)$ is getting larger and larger. We write this as $\lim\limits_{x\to 0^+}f(x)= \infty$

Now we will check what happens from the left of $0$. We will do a similar procedure as above. $\frac{1}{-1}=-1>\frac{1}{-0.1}=-10>\frac{1}{-0.001}=-1000>\ldots>\frac{1}{-0.000001}=-1000000$.

This time we see that as the $x$ values get close and closer to $0$ from the left, $f(x)$ is getting smaller and smaller. We write this as $\lim\limits_{x\to 0^-}f(x)=-\infty$.

We note that $\lim\limits_{x\to 0^-}f(x)\neq \lim\limits_{x\to 0^+}f(x)$, and hence the limit does not exist. If the right sided and left sided limits had been equal, then you could have said that the limit itself was equal.

• Ah okay, that makes sense. How then, without a graph, would you find the limit of a function at a discontinuous point? – OpieDopee Mar 29 '14 at 18:47
• @Ethan: I put in an example into my post. Hopefully it's helpful. – Sujaan Kunalan Mar 29 '14 at 19:12

In the definition of a limit, the value of the function at the point the limit is being taken to is specifically left out.

We say that $\lim_{x\rightarrow a}{f(x)}=L$ if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that $0<|x-a|<\delta$ implies that $|f(x)-L|<\epsilon$. Notice that we ensured that $|x-a|>0$, so that it cannot be possible that $x=a$. As others have pointed out, this shows the sense in which the limit determines what a function acts like at a point based on the nearby points.

For your specific example, the idea is that as $x$ approaches 0, $\sin x$ acts like $x$; for this reason, the limit as $x\rightarrow 0$ is 1.

One way in which you can evaluate it rigorously is using l'hopital's rule: if $\lim_{x\rightarrow a}{f(x)}=\lim_{x\rightarrow a}{g(x)}=0$, then $\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}=\lim_{x\rightarrow a}{\frac{f'(x)}{g'(x)}}$, assuming that the RHS exists and the derivatives $f'(x)$ and $g'(x)$ exist near $x=a$.

Thus, we would find that $$\lim_{x\rightarrow 0}{\frac{\sin x}{x}}=\lim_{x\rightarrow 0}{\cos x}=\cos 0=1.$$

• How do you prove $(\sin x)'= \cos(x)$? – user10444 Mar 29 '14 at 18:26
• By evaluating $\lim_{x\rightarrow 0}{\frac{\sin x}{x}}$ in a different way from the above. But how I would do this depends entirely on how $\sin x$ is defined. I personally take the power series definition, in which case it is entirely obvious. – Hayden Mar 29 '14 at 18:29
• @Hayden: My textbook solves examples such as $\lim_{x\to 0}\frac {tan x}{x}$ by evaluating the function at $x = 0$ . So how could I do a limit without the value of the function at the point the limit is being yaken? – OpieDopee Mar 29 '14 at 18:30
• @Ethan you can only evaluate functions at the point the limit is being taken to if the function is continuous at that point. There are many different methods that can be used to evaluate limits. See here (omega.albany.edu:8008/mat112dir/chap1-3.html) for some of the properties of limits and how they can be applied – Hayden Mar 29 '14 at 18:34
• When you evaluate a limit, you do not just evaluate the function at the value unless you know it is continuous there. In this example clearly $x=0$ is a discontinuity- otherwise you'd be dividing by $0$. – Sujaan Kunalan Mar 29 '14 at 18:34

L'Hopital's rule is an overkill here. If you already know that $\sin'x = \cos x$ (which you need to know to use L'Hopital), then simply by definition of derivatives you have $$\lim_{x \to 0} \frac{\sin x}{x} = \sin'(0) = \cos 0 = 1.$$

If you don't know that $\sin'x = \cos x$, then things start to depend on the definition of $\sin x$. If you define $\sin x$ using the power series $\sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k + 1}}{(2k+1)!}$, then proving $\lim_{x \to 0}\frac{\sin x}{x} = 1$ is not harder than proving that the definition is correct at all, i.e. that the series converges everywhere. If you have a geometric definition of $\sin x$, then the proof will be (at least in part) geometric.

Easy way to view: L'Hospital Rule: $$\lim_{x\rightarrow 0} \frac {\sin x}{x} = \lim_ {x \rightarrow 0} \frac {\cos x}{1} = \cos 0 = 1$$

Another way: first take $0 <\delta <\frac {\pi}{2}$. Thus, $\forall x \in (0, \delta)$ we have the following chain of valid inequalities: $$\sin x \leq x \leq \tan x$$

As $0 <x <\frac {\pi}{2}$, dividing this chain of inequalities for $\sin x \neq$ 0, we obtain $$1 \leq \frac {x}{\sin x} \leq \frac {1}{\cos x}.$$ Taking the inverse, $$\cos x \leq \frac {\sin x}{x} \leq 1$$ Now, for the "sandwich theorem", making the limit as $x \rightarrow 0^+$, we conclude that $$\lim_ {x \rightarrow 0^+} \frac {\sin x}{x} = 1.$$

Similarly it is shown that $$\lim_{x \rightarrow 0^-} \frac{\sin x}{x} = 1.$$

For this, take $x \in (- \delta, 0)$, in this case $\sin x <$ 0, etc..