- #1
zorro
- 1,384
- 0
Homework Statement
I tried expanding e^x and evaluated the limit as 1.
The answer given is 0.
In summary, the conversation is about evaluating the limit of e^x and how the given answer of 0 may not be accurate as it depends on the values of x. The conversation also includes a recommendation to try different values of x and a clarification about a possible typo in the original question. Finally, a suggestion is given to compute the limit by setting x=1/y and examining the limit as y approaches 0.
I tried expanding e^x and evaluated the limit as 1.
The answer given is 0.
- #2
Char. Limit
Gold Member
- 1,222
- 22
How odd. The answer actually depends on the values of x... try a couple of values of x and you'll see what I mean. Might I recommend .75 and 2?
- #3
zorro
- 1,384
- 0
This question is actually a part of another one.
Got any idea now?
- #4
Char. Limit
Gold Member
- 1,222
- 22
Either there's a typo or you misread it. It should be as [tex]x\to\infty[/tex]. Then it makes sense, especially since you're integrating over dx, not over dn.
- #5
zorro
- 1,384
- 0
Oh yes, that must be a typo. I got it now.
Thanks!
- #6
hunt_mat
Homework Helper
- 1,798
- 33
To compute the limit as [tex]x\rightarrow\infty[/tex], set x=1/y amd examine the limit as [tex]y\rightarrow 0[/tex]
FAQ: Evaluating Limit of Expanded e^x: 0?
What is the limit of expanded e^x as x approaches 0?
The limit of expanded e^x as x approaches 0 is equal to 1. This can be shown using the definition of a limit, where the limit of a function at a point is the value that the function approaches as the input approaches that point.
How do you evaluate the limit of expanded e^x at 0?
To evaluate the limit of expanded e^x at 0, you can use the Maclaurin series expansion for e^x, which is 1 + x + x^2/2! + x^3/3! + ... By plugging in 0 for x, you will get the value of 1 for the limit.
Why is the limit of expanded e^x at 0 equal to 1?
The limit of expanded e^x at 0 is equal to 1 because e^x is defined as the infinite sum of x^k/k!, where k goes from 0 to infinity. When x is equal to 0, all the terms in the sum except for the first term (which is 1) become 0. Therefore, the limit becomes 1.
Can the limit of expanded e^x at 0 be evaluated using L'Hospital's rule?
Yes, the limit of expanded e^x at 0 can be evaluated using L'Hospital's rule, which states that the limit of a fraction of two functions can be found by taking the limit of the derivatives of the numerator and denominator. In this case, the limit can be found by taking the limit of the derivative of e^x, which is e^x, and the derivative of x, which is 1. This also results in a limit of 1.
How does the limit of expanded e^x at 0 relate to the derivative of e^x?
The limit of expanded e^x at 0 is equal to the derivative of e^x evaluated at 0. This is because the derivative of e^x is equal to e^x, and when x is equal to 0, the derivative of e^x is also equal to 1. This can be seen from the graph of e^x, where the slope of the tangent line at x=0 is equal to 1.