nowhere ..atleast now ..

yes ..u r right ....thanx...here we have to apply expansion of cosx and ex ..that sure wil do it ..

ans>>

lim x to 0 , [cosx(cosx -1) - e^x( cosx-1) ]/ xⁿ    -x³/2xⁿ

                 =(cosx-1)(e^x-1)/xⁿ   -x³/2xⁿ

                =(2sin²x/2)(e^x-1)/xⁿ   -x³/2xⁿ

           apply 2 of these >

1.  lim x to 0 ,, sinx/x =1

2. lim x to 0 ,, (e^x-1)/x = 1

that simplifies your answer ..

=(1/2)(1)(1)/x^n-3 - 1/2(x^3-n)

=1/2x^3-n - 1/2x^3-n

=0 ..........for n>3

??????????? but you asked for a non zero  ..??  something fishy>>

a^2-b^2 can be written as (a+b)(a-b)

since a^2-b^2 is prime ..either (a+b) or (a-b) will be equal to 1..that too must be noted that the other one should be prime as well ...(bcos a and b are +ve integers)

hence two pts are clear...

1. one among (a+b) or (a-b) is equal to 1.

2. the other is prime...

Now...a and b are +ve integers...only (a-b) can be equalled to 1 ..and not (a+b)..

This does it ...

a^2 -b^2 =(a-b)(a+b)

                =(1)(a+b)

                =(a+b)

nice question ...!!

ask me..

i MISSED it!!

i liked the firefox ..

nice animative avatar.