- #1
Nathew
Homework Statement
Lim x→0 (e^-x+x-1)/(xsinx)
The Attempt at a Solution
I've tried L'Hopital's multiple times but it only gets more complicated.
Limit of (e^-x+x-1)/(xsinx) as x approaches 0
- Thread starter Nathew
- Start date
In summary, the limit of (e^-x+x-1)/(xsinx) as x approaches 0 is equal to 1. This is because as x gets closer and closer to 0, the numerator approaches 0 while the denominator also approaches 0. By applying L'Hopital's rule, we can see that the limit actually approaches 1. This can also be confirmed by graphing the function, which shows that the curve approaches the point (0,1). Thus, the limit of (e^-x+x-1)/(xsinx) as x approaches 0 is equal to 1.
Lim x→0 (e^-x+x-1)/(xsinx)
I've tried L'Hopital's multiple times but it only gets more complicated.
- #2
Ray Vickson
Science Advisor
Homework Helper
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Nathew said:
Show your work.
- #3
tiny-tim
Science Advisor
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Hi Nathew! 
Shouldn't …
Nathew said:
Shouldn't
…show us your first two.
- #4
Mark44
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Thread closed. Nathew, please start a new thread that shows what you have tried.
FAQ: Limit of (e^-x+x-1)/(xsinx) as x approaches 0
1. What does the expression "Lim x→0 (e^-x+x-1)/(xsinx)" represent?
The expression represents the limit of a function as x approaches 0. It is often used to evaluate the behavior of a function at a specific point or to determine the maximum or minimum value of a function.
2. How do you solve for the limit of a function?
To solve for the limit of a function, you need to plug in the value that x is approaching into the function and simplify the expression. If the resulting expression is undefined or indeterminate, you may need to use algebraic techniques or L'Hospital's rule to evaluate the limit.
3. What is the value of "Lim x→0 (e^-x+x-1)/(xsinx)"?
The limit of the expression "Lim x→0 (e^-x+x-1)/(xsinx)" is equal to 1.
4. Can you explain the steps to solve for the limit of this function?
To solve for the limit of this function, you would first plug in 0 for x to get (e^0+0-1)/(0sin0). This simplifies to (1+0-1)/(0*0), which is equal to 0/0. Since this is an indeterminate form, you would then use L'Hospital's rule to evaluate the limit. This involves finding the derivatives of the numerator and denominator, plugging in 0 for x, and then simplifying the resulting expression. In this case, the limit simplifies to (e^0+1)/(sin0+xcos0), which is equal to 1/0. This is still an indeterminate form, so you would use L'Hospital's rule again until you obtain a non-indeterminate form. After two applications of L'Hospital's rule, the limit simplifies to 1.
5. What is the significance of evaluating this limit?
Evaluating this limit can help us understand the behavior of the function near x=0. In this case, the limit tells us that as x approaches 0, the function approaches a value of 1. This can be useful in determining the maximum or minimum value of the function or in analyzing the behavior of the function near x=0.