If \lim\limits_{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17, then 5 a^{2}+b^{2} is equal to

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Answer: B. 68 The Taylor series for e^x, \cos x, and e^{-x} around x=0 are: e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots, \cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots, e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots. Substituting these into the limit and simplifying: \lim\limits_{x ightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}. This simplifies to: \lim\limits _{x ightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}ight) 	imes 4 x^2}. Let's evaluate this expression: (\frac{1-\cos 2x}{(2x)^2}) 	imes 4x^2. We know that 1 - \cos 2x can be rewritten using the double-angle formula for cosine, which is \cos 2x = 1 - 2\sin^2x. So, 1 - \cos 2x = 2\sin^2x. Substituting this into our expression we get: (\frac{2\sin^2x}{(2x)^2}) 	imes 4x^2. This simplifies to: {1 \over 2}(\frac{\sin x}{x})^2 	imes 4x^2 = {1 \over 2} 	imes 4x^2 = 2x^2. So, (\frac{1-\cos 2x}{(2x)^2}) 	imes 4x^2 = 2x^2. 	herefore \lim _{x ightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17. Now, for the limit to exist, the coefficient of x in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation: a-\frac{c}{2} = 0 \Rightarrow c=2a. Then, for the limit to equal 17, the coefficient of x^2 in the numerator must equal 17 	imes 2 = 34. This gives us the equation: \frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68. Since c=2a, we can substitute c in the equation to get: a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.

If $$\lim\limits_{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17$$, then $$5 a^{2}+b^{2}$$ is equal to

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