lim x->0 (e^x-e^sinx)/(x-sinx) solve by l'hospital rule
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Answered by
5
→0
ex
−
e
−x
−2x
x−
sinx
=
e0
−
e
−0
−2
(0)
0−
sin0
=
1−1−0
0−0
=00
This is an indeterminate type so use l'Hopital's Rule which is the limit of the quotient of the derivative of the top and the derivative of the bottom as x goes to 0.
lim
x→0
ex
+
e
−x
−2
1−
cosx
=
e0
+
e
−0
−2
1−
cos0
=
1+1−2
1−1
ex
−
e
−x
−2x
x−
sinx
=
e0
−
e
−0
−2
(0)
0−
sin0
=
1−1−0
0−0
=00
This is an indeterminate type so use l'Hopital's Rule which is the limit of the quotient of the derivative of the top and the derivative of the bottom as x goes to 0.
lim
x→0
ex
+
e
−x
−2
1−
cosx
=
e0
+
e
−0
−2
1−
cos0
=
1+1−2
1−1
Answered by
18
Ans. is 1 ... solution is in the pic....
we have to differentiate it three times ....
we have to differentiate it three times ....
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