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

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3 years ago

```							 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.
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3 years ago
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