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In this case, y= \lim_{x\to 0} f(x) \lim_{x\to 0} g(x) is a meaningless expression. You CAN calculate \lim_{x\to 0} f(x) g(x) but you cannot split it into two limits and give the new expression any meaning.If you take a limit and get 0 * ∞, it means that you're not done yet. The indeterminate form [0 * ∞] is one of several indeterminate forms that can arise when you're taking limits. One thing that is often done is to rewrite the expression so that the limit becomes [0/0] or [∞f

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- #2

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[tex] \lim_{x\to 0} f(x) g(x) [/tex]

but you cannot split it into two limits and give the new expression any meaning.

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This wiki article has a table of indeterminate forms - http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms

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The limit of f(x) = 1/x as x approaches 0 is not zero, it is infinity.

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You're sort of half correct.The limit of f(x) = 1/x as x approaches 0 is not zero, it is infinity.

$$\lim_{x \to 0^+}\frac 1 x = ∞$$

$$\lim_{x \to 0^-}\frac 1 x = -∞$$

Since the one-sided limits aren't equal, the two-sided limit doesn't exist.

- #6

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Sorry for my poor phrasing.

How about in this new example.

y = [limx→0^+ 1/x][limz→0 (1 - cosz)/z]

Would it be possible to evaluate this limit since one limit approaches infinity while the other approaches 0, and they are different variables in this case.

How about in this new example.

y = [limx→0^+ 1/x][limz→0 (1 - cosz)/z]

Would it be possible to evaluate this limit since one limit approaches infinity while the other approaches 0, and they are different variables in this case.

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No, you're right back to what I was talking about in post #3, and what you had in your first post. The fact that you have different variables does not change things.

How about in this new example.

y = [limx→0^+ 1/x][limz→0 (1 - cosz)/z]

Would it be possible to evaluate this limit since one limit approaches infinity while the other approaches 0, and they are different variables in this case.

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How did this limit come up? What's the problem you're trying to solve?

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