Question

plz answer the question step by step​

Answer 1

Step-by-step explanation:

We have,

$f(x) = \bigg \{ { \frac{1 - \cos(4x) }{8 {x}^{2} } ,x \not = 0 \atop{k , \: \: \: \: \: \: \: \: \: \: \: \: \: x = 0 }}$

Since, it is continuous at x=0,

$\lim_{x \rarr0} \frac{1 - \cos(4x) }{8 {x}^{2} } = k$

$\implies \lim_{x \rarr0} \frac{2 \sin^{2} (2x) }{8 {x}^{2} } = k$

$\implies \lim_{x \rarr0} \frac{ \sin^{2} (2x) }{4 {x}^{2} } = k$

$\implies \lim_{2x \rarr0} \frac{ \sin^{2} (2x) }{ {(2x)}^{2} } = k$

$\implies k =1$