# Limits with trig functions

I've been slacking off in calculus and I honestly don't know what to do with this problem. I got that it's in the indeterminate form, but I have no clue where to go from there.

$$\lim_{x \to 0} \frac{\sin(x)}{5x} = \frac{\sin(0)}{0} = \frac{0}{0}$$

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The first thing you ought to know is that you can't replace the value at which the limit is to be evaluated (0, in this case) in the expression. At least not always. – Weltschmerz Aug 30 '11 at 2:08
"in the indeterminate form" - and thus, attacking this with l'Hôpital is so deliciously circular. :) – J. M. Aug 30 '11 at 2:42

This answer consists of two parts, an informal part that uses a calculator, and a more formal part.

What the calculator can tell us: Set your calculator to radian mode. Let $$f(x)=\frac{\sin x}{5x}.$$ Now use the calculator to try to calculate $f(0)$. The calculator will get very upset. Mine shows an E and refuses to do anything until I reset it. That is because our function $f(x)$ is not defined at $x=0$.

In our limit calculation, we want to find what number (if any) is approached by $f(x)$ as $x$ approaches $0$. So let us evaluate $f(x)$ for various $x$ that are near to $0$, but not equal to $0$.

I will start by finding $f(0.1)$. My calculator says that $f(0.1)$ is approximately $0.1996668$. Your calculator, which is undoubtedly fancier, may give an answer to more decimal places.

Now let's calculate $f(-0.05)$. My calculator says that $f(-0.05)$ is approximately $0.1999167$. Next we calculate say $f(0.003)$. My calculator says that $f(0.003)$ is approximately $0.1999997$.

Next we calculate $f(0.0001)$. My calculator says this is approximately $0.2$. Of course, $f(0.0001)$ is not exactly equal to $0.2$, but to the limit of accuracy of the display, that is the value.

If we look at these calculations, it seems sensible to believe that as $x$ approaches $0$, $f(x)$ approaches $0.2$. That is indeed the case.

But you are undoubtedly expected to give some sort of formal argument, not a plausible conjecture based on a few calculations.

A more formal argument: I am sure that your course notes, and your textbook, discuss in detail the key fact that you are expected to use. This key fact is that $$\lim_{x\to 0}\frac{\sin x}{x}=1.$$

The book, and probably the notes, give a justification of this fact. The details depend on exactly how the sine function is defined. In the majority of calculus courses, the justification is geometric. It uses a diagram, and either a comparison of areas or of lengths. Do look at the details. But from the time the key fact is proved, you can use it freely in further calculations.

Note that $$\frac{\sin x}{5x}=\frac{1}{5}\frac{\sin x}{x}.$$ Let $x$ approach $0$. As $x$ approaches $0$, $\frac{\sin x}{x}$ approaches $1$ (this is our key fact). Thus $$\lim_{x\to 0}\frac{\sin x}{5x}=\lim_{x\to 0}\frac{1}{5}\frac{\sin x}{x}=\frac{1}{5}\cdot 1=\frac{1}{5}.$$

Comment: Very informally, when $x$ is close to but not equal to $0$, $\frac{\sin x}{x}$ is roughly $1$, so $\frac{\sin x}{5x}$ is roughly $\frac{1}{5}$. This level of informality is probably not acceptable in your course. And it shouldn't be, for all too often a very informal argument is simply wrong, and gives the wrong answer. However, it can tell you what's really going on. Then you do a more formal verification, as in the write-up above.

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$\lim_{x\to0}{\sin x\over5x}=(1/5)\lim_{x\to0}{\sin x\over x}$, so if you know that second limit, you're done. If you don't know that second limit, there are many nice geometrical evaluations of it on the web. The limit is needed for finding the derivative of $\sin x$, so if you type "derivative" and "sin" into your favorite search engine, lots of webpages with evaluations of the limit will turn up.

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Let $f(x)=\dfrac{\sin x}{x}$. Assume $\lim_{x\rightarrow 0}f(x)$ exists. Then, since $$f(-x)=\frac{\sin \left( -x\right) }{-x}=\frac{\sin x}{x}=f(x),$$ both left and right side limits are equal, i.e. $\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow 0^{-}}f(x)$. Hence it is enough to consider the angle $x$ (measured in radians) located in the first quadrant of the trigonometric circle, where the following double inequality is valid (see sketch)

$$\sin x<x<\tan x,\qquad x\in ]0,\frac{\pi }{2}[.$$

Dividing by $\sin x$, we get

$$1<\frac{x}{\sin x}=\frac{1}{f(x)}<\frac{1}{\cos x}.$$ But, since $\displaystyle\lim_{x\rightarrow 0}\dfrac{1}{\cos x}=1$, by the squeeze theorem $\displaystyle\lim_{x\rightarrow 0}\frac{x}{\sin x}=\lim_{x\rightarrow 0}\dfrac{1}{f(x)}=1$. By the algebraic rules of the limit of a function $\lim_{x\rightarrow 0}f(x)$ exists and its value is

$$\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,$$

and the requested limit is

$$\lim_{x\rightarrow 0}\frac{\sin x}{5x}=\frac{1}{5}\lim_{x\rightarrow 0}\frac{\sin x}{x}=\frac{1}{5}.$$

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