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
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
3 answers
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
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 :)
1 answer