If lim _{x rightarrow 0} frac{e^{a x}-cos (b | vedclass

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(C) Given $\lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17$.Using the expansion $e^{ax} = 1 + ax + \frac{a^2

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$68$

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(C) Given $\lim _{x ightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17$.Using the expansion $e^{ax} = 1 + ax + \frac{a^2x^2}{2} + \dots$,$\cos(bx) = 1 - \frac{b^2x^2}{2} + \dots$,$e^{-cx} = 1 - cx + \dots$,and $1 - \cos(2x) = 2x^2 + \dots$ as $x 	o 0$:$\lim _{x ightarrow 0} \frac{(1 + ax + \frac{a^2x^2}{2}) - (1 - \frac{b^2x^2}{2}) - \frac{cx}{2}(1 - cx)}{2x^2} = 17$$\lim _{x ightarrow 0} \frac{(a - \frac{c}{2})x + x^2(\frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2})}{2x^2} = 17$For the limit to exist,the coefficient of $x$ must be zero:$a - \frac{c}{2} = 0 \implies c = 2a$Substituting $c = 2a$ into the limit expression:$\frac{\frac{a^2}{2} + \frac{b^2}{2} + \frac{(2a)^2}{2}}{2} = 17$$\frac{a^2 + b^2 + 4a^2}{4} = 17$$5a^2 + b^2 = 68$

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

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