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Limits, Continuity and Differentiability Part 3

Section: Mathematics

Question : 77 of 79

Marks: +1, -0

\lim\limits_{x \to 0} \frac{ e^{a x} - \cos (b x) - \frac{c x e^{-c x}}{2} }{ 1 - \cos (2 x) } = 17

5 a^2 + b^2

Solution:

\lim\limits_{x \to 0} \frac{ e^{a x} - \cos b x - \frac{c x e^{-c x}}{2} }{ 1 - \cos 2 x } = 17

On expansion

\lim\limits_{x \to 0} \frac{ \left(1 + a x + \frac{(a x)^2}{2!} + \dots\right) - \left(1 - \frac{(b x)^2}{2!} + \dots\right) - \frac{c x}{2} \left(1 - c x + \frac{(c x)^2}{2!}\right) }{ \left( \frac{1 - \cos 2 x}{(2 x)^2} \right) \times (2 x)^2 } = 17

\lim\limits_{x \to 0} \frac{ x \left(a - \frac{c}{2}\right) + x^2 \left( \frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2} \right) }{ \frac{1}{2} (4 x^2) } = 17

For limit to be exist

a - \frac{c}{2} = 0 \Rightarrow c = 2 a

\Rightarrow \frac{ \frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2} }{2} = 17

\Rightarrow \frac{a^2}{2} + \frac{b^2}{2} + \frac{4 a^2}{2} = 34

\Rightarrow 5 a^2 + b^2 = 68

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