lim x tends to 0 (e^sinx-x-x^2)/x^3
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Hlw mate!!
Answer:
2
Explanation:
limx→0ex−e−x−2xx−sinx=e0−e−0−2(0)0−sin0=1−1−00−0=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.
limx→0ex+e−x−21−cosx=e0+e−0−21−cos0=1+1−21−1=00
Still an indeterminate form so apply l'Hopital's rule again
limx→0ex−e−xsinx=e0−e−0sin0=1−10=00
Still an indeterminate form so apply l'Hopital's rule again
limx→0ex+e−xcosx=e0+e−0cos0=1+11=2
Hope it helpful
Answer:
2
Explanation:
limx→0ex−e−x−2xx−sinx=e0−e−0−2(0)0−sin0=1−1−00−0=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.
limx→0ex+e−x−21−cosx=e0+e−0−21−cos0=1+1−21−1=00
Still an indeterminate form so apply l'Hopital's rule again
limx→0ex−e−xsinx=e0−e−0sin0=1−10=00
Still an indeterminate form so apply l'Hopital's rule again
limx→0ex+e−xcosx=e0+e−0cos0=1+11=2
Hope it helpful
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