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Lim x tends to zero ae^x-bcosx+ce^-x/xsinx=2 find the value of a b and c
lim (x→0) (aex - b cos x + ce(-x))/(x sin x) = 2Let f(x) = aex - b cos x + ce- xAnd g(x) = x sin xlim (x→0) g(x) = 0lim (x→0) (aex -b cos x + ce(-x))/(x sin x) = 2So, f(x) should be zero for finite limit of f(x)/g(x)⇒ a + c = b ....... (i)⇒ L = lim (x→0) (aex - b cos x + ce(-x))/(x sin x) [0/0 form]Again using L' Hospital's Rule= lim (x→0) (aex + b sin x - ce(-x))/(x cos x + sin x)Denominator = lim (x→0) x cos x + sin x = 0So for finite, lim (x→0) aex + b sin x – ce(-x) = 0⇒ a - c = 0a = c ........ (ii)L = lim (x→0) (aex + (2a) sin x – ae(-x))/(x cos x + sin x ) [0/0 form] (using (i) and (ii)) = lim (x→0) (a(ex + e(-x)) + 2a cos x)/(-x sin x + 2 cos x) (applying L' Hospital rule) = lim (x→0) (a(ex + e(-x)) + 2a cos x)/(-x sin x + 2 cos x) = (2a + 2a)/2 = 2aGiven the value of limit is 2Hence 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.
lim (x→0) (aex - b cos x + ce(-x))/(x sin x) = 2
Let f(x) = aex - b cos x + ce- x
And g(x) = x sin x
lim (x→0) g(x) = 0
lim (x→0) (aex -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 (x→0) (aex - b cos x + ce(-x))/(x sin x) [0/0 form]
Again using L' Hospital's Rule
= lim (x→0) (aex + b sin x - ce(-x))/(x cos x + sin x)
Denominator = lim (x→0) x cos x + sin x = 0
So for finite, lim (x→0) aex + b sin x – ce(-x) = 0
⇒ a - c = 0
a = c ........ (ii)
L = lim (x→0) (aex + (2a) sin x – ae(-x))/(x cos x + sin x ) [0/0 form] (using (i) and (ii))
= lim (x→0) (a(ex + e(-x)) + 2a cos x)/(-x sin x + 2 cos x) (applying L' Hospital rule)
= lim (x→0) (a(ex + 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.
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