Lf the function fx-1-cos 4xx2 if x 0is continuous at x-0 then a-

The correct option is C 8 Given f(x) is continuous at x=0. lim _ x→ 0 f(x)=f(0) Here , f(0)=a =lim _ x→ 0 ^ 1 cos 4x nbsp;/ x ^ 2 ...

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Definite Integral as Limit of Sum

lf the functi...

Question

lf the function $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}} & i f & x < 0 a & i f & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} & i f & x > 0 \end{matrix}$
is continuous at $x = 0$ then $a =$

8

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a

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & x < 0 a, & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{matrix}

x = 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & w h e n x < 0 a, & w h e n x = 0 \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}, & w h e n x > 0 \end{matrix}

x = 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & i f & x < 0 a, & i f & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{4}} - 4}, & i f & x > 0 \end{matrix}

a, f

x = 0

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{{sin}^{2} a x}{x^{2}}, x \neq 0 1, x = 0 \end{matrix}

x = 0

a =

k

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{1 - cos 4 x}{8 x^{2}} & , x \neq 0 k & , x = 0 \end{matrix}

x = 0

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