1. ## limit question

i would think the lim as x-->0 of f(x)/f(x) would be 1 regardless of what f(x) is. i am looking at a solution in the back of a calculus book and finding a solution of "limit doesn't exist." the function f(x) is sin(1/x). am i missing something?

2. In considering the function $\displaystyle f(x) = \sin \left( {\frac{1}{x}} \right)$, in any neighborhood of 0 $\displaystyle f(x)$ has infinitely many zeros.
Therefore, because $\displaystyle \frac{0}{0}$ is undefined so is $\displaystyle \frac{{f(\lambda )}}{{f(\lambda )}}$ where $\displaystyle \lambda$ is a zero of $\displaystyle f(x)$.

Because there are infinitely many places where $\displaystyle \frac{{f(x )}}{{f(x )}}$ is undefined is any neighborhood of 0, $\displaystyle \lim _{x \to 0} \frac{{f(x)}}{{f(x)}}$ does not exist.

3. To speak of the limit,
$\displaystyle \lim_{x\to 0} g(x)$

We need to be sure that $\displaystyle g(x)$ is defined on some open interval containing $\displaystyle 0$ except possibly at $\displaystyle 0$ it self.

So if $\displaystyle f(x) = \sin \frac{1}{x}$.
And you define,
$\displaystyle g(x) = \frac{f(x)}{f(x)}$
You cannot talk about the limit,
$\displaystyle \lim_{x\to 0}g(x)$
Because as Plato says $\displaystyle g(x)$ is not defined in some "neighborhood".

4. thanks!

5. Originally Posted by ThePerfectHacker
$\displaystyle g(x) = \frac{f(x)}{f(x)}$
You cannot talk about the limit,
$\displaystyle \lim_{x\to 0}g(x)$
Because as Plato says $\displaystyle g(x)$ is not defined in some "neighborhood".
Please note: $\displaystyle g(x)$ is not defined at some point in any neighborhood of 0.

6. noted, thanks!