My question is:Use L'Hospital's rule to find limit as x approaches 0 (e^x-e^-x-2x)/x-sinx. 1st I took the derivative to get (e^x-e^-x-2)/1+cosx. Substituting in 0, I got -1. The answer is supposed to be 2.I also tried (e^x-xe^-x-2)/1+cosx, and wound up with -(1/2). I think I might be supposed to be using a hyberbolic identity, but I can't figure out how. Any help is appreciated
watch out for sign errors !
derivative of top = (e^x + e^-x -2)
when x = 0 this is 2-2 = 0
derivative of bottom = (1-cos x)
when x = 0 this is 1-1 = 0
well that was not a big help so try next derivative
top: (e^x - e^-x )
oh my, zero again
bottom: sin x
also 0 again
third time
top: (e^x + e^-x)
---> 2 whew !
bottom: cos x
----> 1
2/1 = 2
You can use L'Hospital rule several times as long as the form 0/0 or infinity/infinity is there.
Using sinh(x) = 1/2(e^2 - e^-2); limit is
2sinh(x)-2x/x-sin(x) which is 0/0 form.
applying L'Hospital rule; limit is
2(cosh(x)-1)/1-cos(x) which is 0/0 form.
apply L'Hospital two times, you'll get
2(cosh(x)/cos(x)=2 when x to 0
That works too :)
To find the limit as x approaches 0 of the expression: (e^x - e^(-x) - 2x) / (x - sin(x)), you correctly started by taking the derivative of the numerator and denominator using L'Hôpital's rule. However, there seems to be an error in your derivative calculation.
Let's go through the correct steps:
1. Start with the original expression:
(e^x - e^(-x) - 2x) / (x - sin(x))
2. Take the derivative of the numerator and denominator separately:
Numerator:
[d/dx] (e^x - e^(-x) - 2x) = e^x + e^(-x) - 2
Denominator:
[d/dx] (x - sin(x)) = 1 - cos(x)
Now, we have the expression:
(e^x + e^(-x) - 2) / (1 - cos(x))
3. Substitute x = 0 into the derived expression:
(e^0 + e^(0) - 2) / (1 - cos(0))
= (1 + 1 - 2) / (1 - 1)
= 0/0
Note: We still have an indeterminate form of 0/0, meaning further simplification is required.
4. We can simplify the expression further using the hyperbolic identity, which states that:
(e^x + e^(-x)) / 2 = cosh(x)
Rearrange the original expression using this identity:
[(e^x + e^(-x)) - 2] / (1 - cos(x))
= 2 * [(e^x + e^(-x)) - 2] / (2 * (1 - cos(x)))
= 2 * [(e^x + e^(-x)) - 2] / 2 * (1 - cos(x))
Now, let's substitute x = 0 into the rearranged expression:
[2 * [(e^0 + e^(0)) - 2]] / [2 * (1 - cos(0))]
= [2 * (1 + 1 - 2)] / (2 * (1 - 1))
= [2 * 0] / (2 * 0)
= 0/0
5. We still have an indeterminate form 0/0, so let's apply L'Hôpital's rule again:
Take the derivative of the numerator and denominator:
Numerator:
[d/dx] [(e^x + e^(-x)) - 2]
= (e^x - e^(-x))
Denominator:
[d/dx] [2 * (1 - cos(x))]
= 2 * sin(x)
Substitute x = 0 into the derived expression:
[(e^0 - e^(0))] / [2 * sin(0)]
= (1 - 1) / (2 * 0)
= 0/0
6. We still have an indeterminate form 0/0, so let's apply L'Hôpital's rule one more time:
Numerator:
[d/dx] (e^x - e^(-x))
= e^x + e^(-x)
Denominator:
[d/dx] (2 * sin(x))
= 2 * cos(x)
Substitute x = 0 into the derived expression:
[e^0 + e^(0)] / [2 * cos(0)]
= (1 + 1) / (2 * 1)
= 2/2
= 1
Therefore, the limit as x approaches 0 of the given expression is 1, not 2 as you mentioned. It seems there might have been a calculation error in your work.