Lim e^x-e^-x-2x

x-sinx As x tends to zero

a)1

b)-1

c)2

d)0

Plz explain.......

apply l-hospital's rule

e^x+e^-x-2/(1-cosx)

again apply l-hospital's rule

e^x-e^-x/sinx 

again apply l-hospital's rule

e^x+e^-x/cosx

put the limit x tend to zero

we get answer as 2

Answer:2

Since it is 0/0 on putting x=0 apply L'Hospital rule  ie  differentiate numerator and denominator with respect to x

Hence we get lim x(tends to 0) (e^x + e^-x-2)/1-cosx

Again it is 0/0

hence differentiating again we get

lim (e^x-e^-x)/sinx

Again it is 0/0

so apllying l hospital rule

lim (e^x+e^-x)/cosx

Now put x=0

we get lim = 2