Lim e x -e -x -2x x-sinx As x tends to zero a)1 b)-1 c)2 d)0 Plz explain.......

Question

Krishna Kant Tiwari · Answer

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

abhishek ganguly · Answer

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

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Lim e x -e -x -2x x-sinx As x tends to zero a)1 b)-1 c)2 d)0 Plz explain.......

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

x-sinx As x tends to zero

a)1

b)-1

c)2

d)0

Plz explain.......

2 Answers

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Lim e x -e -x -2x x-sinx As x tends to zero a)1 b)-1 c)2 d)0 Plz explain.......

Lim ex-e-x-2x x-sinx As x tends to zero a)1 b)-1 c)2 d)0 Plz explain.......

2 Answers

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

x-sinx As x tends to zero

a)1

b)-1

c)2

d)0

Plz explain.......