The discussion centers on the evaluation of limits involving the indeterminate form 0 multiplied by infinity, specifically in the context of the functions f(x) = 1/x and g(z) = (1 - cos(z))/z. It is established that while the limit of f(x) approaches infinity as x approaches 0, the limit of g(z) approaches 0. The multiplication of these limits is deemed meaningless without further manipulation, as one cannot simply multiply the limits of two functions that yield indeterminate forms. The discussion emphasizes the necessity of rewriting such expressions to forms like 0/0 or ∞/∞ to apply L'Hôpital's Rule for evaluation.
PREREQUISITESStudents and educators in calculus, mathematicians dealing with limits, and anyone seeking to deepen their understanding of indeterminate forms in mathematical analysis.
MathewsMD said:If you have f(x) = 1/x and g(a) = (cosx - 1)/x and then y = [limx→0 f(x)][limx→0g(x)], the two individual limits equal 0 and infinity, respectively. Since these are limits and only approach these values, would the multiplication of the two limits equal 0, infinity or something else?
You're sort of half correct.SteamKing said:The limit of f(x) = 1/x as x approaches 0 is not zero, it is infinity.
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.MathewsMD said: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.