# Finding a value so that a given f(x) is continuous at 0

I need help in the following question:

Let $$f(x) = \left\{\begin{array}{ll} \frac{\sin (0.19x)}{x} &\text{when }x\neq 0\\ L&\text{when }x=0 \end{array}\right.$$

Find the value of $L$ so that $f(x)$ is continuous at 0.

Thanks.

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Please note: the backslash to indicate division is ambiguous without parenthesis. When you write sin 0.19x/x, the standard interpretation is that the division goes before the sine, so you would be telling us that you are looking at $$\sin\left(\frac {0.19x}{x}\right)$$which is probably *not* what you are actually looking at. You would want to write (sin 0.19x)/x to describe $$\frac{\sin(0.19x)}{x}.$$But it will be much better if you learn a little bit of $\LaTeX$ and mark-up your posts. Thank you. –  Arturo Magidin Mar 14 '12 at 18:34

By definition, a function $f(x)$ is continuous at $a$ if and only if:

1. $f(x)$ is defined at $x=a$;
2. $\lim\limits_{x\to a}f(x)$ exists; and
3. $\lim\limits_{x\to a}f(x) = f(a)$.

So for your $f(x)$ to be continuous at $0$ you need it to be defined at $0$ (which it is), and you need $$L = f(0) = \lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}\frac{\sin(0.19x)}{x}$$ to be true.

Since you are free to decide what you want $L$ to be, what you need to do is figure out how much that limit is.

HINT: you've probably recently seen the fact that $\lim\limits_{u\to 0}\frac{\sin u}{u} = 1$. Try to use that.

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Hi Arturo, I understand the concept of continuity but when I approach to compute f(0) of f(x), I encounter the problem of the denominator being 0, which makes the function undefined. Can you enlighten me on how I should approach this issue? Yes I have probably guessed that I have to use the hint that you have shown. –  Xavier Mar 14 '12 at 18:41
@Xavier: You don't encounter problems with the denominator being equal to $0$, because when you take a limit as $x$ approaches $0$, you never actually evaluate at $x=0$; that's the one point where you don't care what happens (you only care what happens near $0$, not at $0$). You want to manipulate the limit using valid limit laws so that it looks like the limit in question, which you already know how to evaluate. For instance, would you be able to do the limit if it was $$\lim_{x\to 0}\frac{\sin(0.19x)}{0.19x}$$instead? If so, can you transform the limit so that it looks like that? –  Arturo Magidin Mar 14 '12 at 18:58
Thanks Arturo, I get it now. In this case, my answer should be L = 0.19 –  Xavier Mar 14 '12 at 19:02

One definition of continuity at a point $p$ is that the value of a function at $p$ is equal to the limit of the function as it approaches $p$. Thus what you want to do is set $$L=f(0)=\lim\limits_{x\to 0}\frac{\sin(.19x)}{x}$$ and I will let you compute that limit yourself.

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Thanks for the feedback Alex. –  Xavier Mar 14 '12 at 18:38