Exploring the Limit of -e^(-infinity) in Mathematics

  • Thread starter gillgill
  • Start date
In summary, the value of -e^(-infinity) is equal to 0 and cannot be simplified to a different value. It does have a limit of 0 as the input approaches infinity. The graph of -e^x approaches -e^(-infinity) from the positive side on the y-axis. And finally, -e^(-infinity) is an indeterminate form that cannot be evaluated without knowing the context of the function and its limit.
  • #1
gillgill
128
0
does -e^(-infinity) equals 0?
 
Physics news on Phys.org
  • #2
The limit of e^x as x-->-infinity is 0.
 
  • #3
Are you referring to [itex]\lim_{x\rightarrow\infty} -e^{-x}[/itex] ?
 
  • #4
yes...
 
  • #5
As kamataat said, [itex]\lim_{x\rightarrow\infty} e^{-x}= 0[/itex] and so
[itex]\lim_{x\rightarrow\infty} -e^{-x}= -0= 0[/itex].
 
  • #6
i see. thanks.
 

FAQ: Exploring the Limit of -e^(-infinity) in Mathematics

1. What is the value of -e^(-infinity)?

The value of -e^(-infinity) is equal to 0.

2. Can -e^(-infinity) be simplified to a different value?

No, -e^(-infinity) cannot be simplified to a different value as it approaches 0 in the limit as infinity increases.

3. Does -e^(-infinity) have a limit?

Yes, as the input of -e^x approaches infinity, the output approaches 0, giving -e^(-infinity) a limit of 0.

4. How does the graph of -e^x approach -e^(-infinity)?

The graph of -e^x approaches 0 as x approaches infinity, meaning it approaches -e^(-infinity) from the positive side on the y-axis.

5. Is -e^(-infinity) an indeterminate form?

Yes, -e^(-infinity) is an indeterminate form as it cannot be evaluated without knowing the context of the function and its limit.

Similar threads

Replies
10
Views
1K
Replies
14
Views
440
Replies
1
Views
535
Replies
10
Views
782
Replies
13
Views
1K
Replies
23
Views
1K
Back
Top