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

AI Thread Summary
In calculus, when evaluating limits, a quotient f(x)/g(x) where f(c) = 0 and g(c) = 0 results in an indeterminate form, but this does not guarantee the existence of like factors in both the numerator and denominator that can be canceled. The example f(x) = (x^2 - 1)/(x - 1) illustrates that while 0/0 can be resolved by factoring, not all functions behave similarly. The function |x|/x also approaches 0/0 as x approaches 0, yet its limit does not exist, demonstrating that indeterminate forms do not imply factorability. A counterexample provided is sin(x)/x, which yields 0/0 at x=0 but has a limit of 1. The discussion emphasizes the need for caution in assuming that indeterminate forms always indicate cancelable factors.
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 forf(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.
 
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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.
 
CRGreathouse said:
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.
 
Daniel Y. said:
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.
 
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