First slide

Introduction to limits

Question

$if \underset{x \to 0}{L t} \frac{a \cos x + b x \sin x - 5}{x^{4}} is finite, then a =$

Moderate

A

$\frac{5}{2}$
B

5
C

$\frac{2}{5}$
D

$\frac{1}{5}$

Solution

$\underset{x \to 0}{L t} \frac{a \cos x + b x \sin x - 5}{x^{4}}$
= $\lim_{x \to 0} \frac{a (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - . .) + b x (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . .) - 5}{x^{4}}$
$\lim_{x \to 0} \frac{a + x^{2} (b - \frac{a}{2}) + x^{4} (\frac{a}{4!} - \frac{b}{3!}) - 5}{x^{4}} = f i n i t e \Rightarrow a - 5 = 0, b - \frac{a}{2} = 0, \frac{a}{4!} - \frac{b}{3!} = l i m i t v a l u e$

$L i m i t e x i s t s \Rightarrow a - 5 = 0 \Rightarrow a = 5.$

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