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

[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 canceled 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 canceled out? If so how is this proven? Even a simple why would be much appreciated.

 

[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.

 

[Daniel Y.]

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.

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.

 

[rock.freak667]

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.

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


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