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Lim ex-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
ex+e-x-2/(1-cosx)
again apply l-hospital's rule
ex-e-x/sinx
ex+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) (ex + e-x-2)/1-cosx
Again it is 0/0
hence differentiating again we get
lim (ex-e-x)/sinx
so apllying l hospital rule
lim (ex+e-x)/cosx
Now put x=0
we get lim = 2
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