# Lim x tends to zero ae^x-bcosx+ce^-x/xsinx=2 find the value of a b and c

Arun
25750 Points
6 years ago

lim (x0) (ae- b cos x + ce(-x))/(x sin x) = 2

Let f(x) = ae- b cos x + ce- x

And g(x) = x sin x

lim (x0) g(x) = 0

lim (x0) (ae-b cos x + ce(-x))/(x sin x) = 2

So, f(x) should be zero for finite limit of f(x)/g(x)

⇒ a + c = b ....... (i)

⇒ L = lim (x0) (ae- b cos x + ce(-x))/(x sin x) [0/0 form]

Again using L' Hospital's Rule

= lim (x0) (aex + b sin x - ce(-x))/(x cos x + sin x)

Denominator = lim (x0) x cos x + sin x = 0

So for finite, lim (x0) aex + b sin x – ce(-x) = 0

⇒ a - c = 0

a = c ........ (ii)

L = lim (x0) (ae+ (2a) sin x – ae(-x))/(x cos x + sin x ) [0/0 form] (using (i) and (ii))

= lim (x0) (a(e+ e(-x)) + 2a cos x)/(-x sin x + 2 cos x) (applying L' Hospital rule)

= lim (x0) (a(e+ e(-x)) + 2a cos x)/(-x sin x + 2 cos x)

= (2a + 2a)/2

= 2a

Given the value of limit is 2

Hence 2a = 2

⇒ a = 1

⇒ c = a = 1 (from equation (ii))

⇒ b = a + c = 1 + 1 = 2 (from equation (i))

⇒ a = 1, b = 2, c = 1.