$\frac{a c o s x + b x s i n x - 5}{x^{4}}$ is finite, then $a =$

\frac{5}{2}

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5

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\frac{2}{5}

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\frac{1}{5}

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Solution

The correct option is B $5$
Given
${lim}_{x \to 0} \frac{a c o s x + b x s i n x - 5}{x^{4}}$
$= {lim}_{x \to 0} \frac{a (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!}) + b x (x - \frac{x^{3}}{3!}) - 5}{x^{4}}$
$= {lim}_{x \to 0} \frac{(a - 5) + x^{2} (b - \frac{a}{2}) + x^{4} (\frac{a}{4!} - \frac{b}{3!})}{x^{4}}$
Limit is infinite
when $a - 5 = 0, b - \frac{a}{2} = 0$
$∴ a = 5$

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