# The existence of like factors in num/denom if in indeterminate form?

1. Jun 27, 2008

### Daniel Y.

So I'm studying infinite limits in my calculus text (seemed close enough to good old arithematic to put in general math, though), and the following rule is mentioned:

Given two functions f(x) and g(x) defined for all real numbers, when given the quotient $$f(x)/g(x)$$ where f(c) is not 0 and g(c) is 0, there is a vertical asymptote at c. But for$$f(c)/g(c)$$ where f(c) = 0 and g(c) = 0 it is not guarenteed that there is a vertical asymptote at c.

Now obviously the crappy off-the-top-of-my-head definition isn't the perfect one given in the book, but I'm sure if you're able to help you know the one I'm talking about. Now here's the thing:

Every time I've done an exercise that was in indeterminate form for a value c and found analytically for a value of x that isn't defined, I've found that the numerator and denominator have the same factor that can be cancelled out. Consider the following:

$$f(x) = (x^2 -1)/(x-1)$$, when you 'input' f(1) you get 0/0, indeterminate form. But you can factor the equation to $$(x+1)(x-1)/(x-1)$$, cancel out the x-1, and see that f(1) is really 2 (or, at least, a function agreeing at every point except x = 1 is really 2).

This has been my experience with all exercises involving indeterminate form. So my question(s) becomes: if $$f(c)/g(c) = 0/0$$, then does this imply the existence of like factors in both the numerator and denominator that can be factored out and cancelled out? If so how is this proven? Even a simple why would be much appreciated.

Last edited by a moderator: Jun 27, 2008
2. Jun 27, 2008

### CRGreathouse

It does not imply that. $$|x|/x$$ tends to the indeterminate 0/0 as x -> 0, but its limit does not exist.

Also, more broadly, not all functions have easily-expressible form.

3. Jun 27, 2008

### Daniel Y.

Could you give me an example of a function which tends to indeterminate form for some c, and has a limit like in my example, but doesn't have the factorable feature I mentioned in the OP? Thanks for the elucidation CR, but I need my hunch proven wrong with those conditions or it'll still be nagging at me (I can't think of any examples where it doesn't work, myself). Thanks.

4. Jun 27, 2008

### rock.freak667

$$\frac{sinx}{x}$$

if you put x=0 you'll get 0/0 but the limit is actually 1.