If lim_x→0 [ (e^(a−1)x + 2cosbx + (c−2)e^−x) / (xcosx − log_e(1+x)) ] = 2, find a² + b² + c².

Solution:

Using standard expansions:

e^(a−1)x = 1 + (a−1)x + ...

2cosbx = 2 − b²x² + ...

(c−2)e^−x = (c−2)(1 − x + ...)

Denominator:

xcosx − log(1+x)

≈ x(1) − (x − x²/2)

= x²/2

For limit to exist, constant term and x term in numerator must be zero.

Constant term:

1 + 2 + (c−2) = 0

c + 1 = 0

c = −1

Coefficient of x:

(a−1) + (−c+2) = 0

(a−1) + 3 = 0

a = −2

Now comparing x² terms gives:

2(1 − b²) = 2

b = 0

Therefore,

a² + b² + c²

= (−2)² + 0² + (−1)²

= 4 + 1

= 5

Answer: 5