Question

find by applying L Hospital rule​

Answer 1

Answer:

1

Step-by-step explanation:

lim x->0  (e^x-e^-x-2log(1+x))/x sinx

diff. num and den.

lim x->0 (e^x+e^-x -2/1+x) / xcosx+sinx

it's still in 0/0 form

again diff. num. and den.

lim x->0 (e^x-e^-x +2/(1+x)^2 ) / cosx - xsinx + cos x

now put x=0

lim x->0 (e^0-e^0+2/(1+0)^2) / cos 0 -0 +cos 0

 =2/2=1

Answer 2

Answer:

Step-by-step explanation:

dividing numerator and denominator by x

we get

lim x tends 0 (e^x-e^-x-2log(1+x))/x / xsinx/x

we know lim x tends 0 sin x /x =1

lim x tends 0 ln(1+x)/x =1

final expression you get is

lim x tends to 0 (e^x/x)-(e^-x/x)-2 /x

apply l hospital rule

you get

lim x tends to 0 e^x-e^-x

=1-1

=0