Indeterminate: 0 times infinity, limits

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SUMMARY

The discussion centers on the evaluation of limits for three functions as x approaches 2, specifically addressing the indeterminate form of 0 times infinity. The functions are defined as f(x)=5x(x-2)(1/(x-2)), g(x)=5x(x-2)(1/(x-2)^2), and h(x)=5x(x-2)^2(1/(x-2). The limits calculated are 10 for f(x), infinity for g(x), and 0 for h(x). The conversation clarifies that the indeterminate form arises from the behavior of the function as it approaches the limit, rather than at the point itself, highlighting the concept of removable discontinuity.

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I have this problem, with three functions:

f(x)=5x(x-2)(1/(x-2))
g(x)=5x(x-2)(1/(x-2)^2)
h(x) 5x(x-2)^2(1/(x-2)

The problem says that as x approaches 2, the functions take the form 0 times infinity, and asks you to "show" this. I tried plugging in 2 for x, but I got 0 times 1/0 (either undefined by 1/0 or indeterminate by 0/0) in each case. How am I supposed to get 0 times infinity?

The next part of the problem asks for the limits as x approaches 2, which I can see are 10, infinity (magnitude increasing without bound) and 0, respectively. The part after that says "Describe three things that the indeterminate form 0 times infinity could approach." Does that refer to the limits I just found, or some mathy smart thing I don't know about?

Your help would be much appreciated! Thanks!
 
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When x approaches 2, the equation 1/(x-2) gets infinitely large. This doesn't mean when it reaches 2, at which point there is a holy in the graph (removable discontinuity), but when it approaches it. If you were to plug in 1.999999999999999999, you would see the number achieve great size.
So, generally speaking, when it comes to limits, if you obtain 1/0, it means infinity.
 
Oh! That makes sense.. thank you soo much for your help!
 

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