Indeterminate: 0 times infinity, limits

In summary, the problem is asking to show that as x approaches 2, the three given functions take the form of 0 times infinity. The limits of the functions as x approaches 2 are 10, infinity, and 0 respectively. The next part of the problem asks to describe three things that the indeterminate form 0 times infinity could approach. This refers to the limits found previously. When x approaches 2, the equation 1/(x-2) gets infinitely large, but this does not mean that the function reaches 2. When dealing with limits, obtaining 1/0 represents infinity.
  • #1
Kamo
2
0
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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  • #2
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.
 
  • #3
Oh! That makes sense.. thank you soo much for your help!
 

FAQ: Indeterminate: 0 times infinity, limits

What does it mean for a limit to be indeterminate?

An indeterminate limit is one where the value cannot be determined by simply plugging in values for the variable. This usually occurs when the limit approaches infinity or when there is a division by zero. In these cases, further analysis or techniques, such as L'Hopital's rule, are needed to find the limit.

Is indeterminate: 0 times infinity always equal to 0?

No, indeterminate: 0 times infinity can have various values depending on the specific limit being evaluated. This type of limit can approach zero, infinity, or any real number, depending on the behavior of the function.

How do you evaluate a limit that is indeterminate: 0 times infinity?

One way to evaluate an indeterminate: 0 times infinity limit is by using L'Hopital's rule, which involves taking the derivative of the numerator and denominator separately and then evaluating the limit again. Another method is to rewrite the limit in a different form, such as using algebraic manipulation or factoring.

Can an indeterminate limit be evaluated using direct substitution?

No, an indeterminate limit cannot be evaluated using direct substitution because it will result in an undefined expression, such as 0/0 or ∞/∞. Direct substitution is only valid for limits that are not indeterminate, where the value can be determined by simply plugging in values for the variable.

Are there any special cases where an indeterminate: 0 times infinity limit can be evaluated without using special techniques?

Yes, there are certain cases where an indeterminate: 0 times infinity limit can be evaluated without using special techniques. For example, if the function can be simplified or factored to cancel out the indeterminate form, then the limit can be evaluated using direct substitution. Another case is when the limit is a definite integral, where techniques such as the fundamental theorem of calculus can be used to evaluate the limit.

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