1
$\begingroup$

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.

Any comments or advice will be much appreciated.

Thanks.

  • 1
    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.2012-03-14

2 Answers 2

2

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.

  • 0
    Thanks Arturo, I get it now. In this case, my answer should be L = 0.192012-03-14
1

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.

  • 0
    Thanks for the feedback Alex.2012-03-14