Question

if `lim_(x→0) (ae^x - bcosx + ce^-x)/(xsinx)` = 2, then a + b + c is equal to ______.

विकल्प

• 1

• 2

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• 4

Answer 1

if `lim_(x→0) (ae^x - bcosx + ce^-x)/(xsinx)` = 2, then a + b + c is equal to 4.

Explanation:

Given that `lim_(x→0) (ae^x - bcosx + ce^-x)/(xsinx)` = 2  ...(i)

Since, limit exists, therefore one of the indeterminant form will be present.

Now for indeterminant `(0/0)` form

Since, e^x = `1 + x + x^2/(2!) + ...`, cosx = `1 - x^2/(2!) + x^4/(4!) ...`,  e^–x = `1 - x + x^2/(2!)....`

Using equation (i)

`lim_(x→0) (a(1 + x + x^2/(2!) + ...)-b(1 - x^2/(2!) + x^4/(4!) - ...) + c(1 - x + x^2/(2!) - ....))/(x(sinx/x)(x))` = 2

`lim_(x→0) ((a - b + c) + x(a - c) + x^2(a/2 + b/2 + c/2))/(x^2(sinx/x))` = 2

Comparing LHS and RHS

(a – b + c) = 0, (a – c) = 0, `(a + b + c)/2` = 2

⇒ a + b + c = 4