Question

Find k if
$f(x) = \frac{1 - \cos(4x) }{ {x}^{2} } \: x < 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: k \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:x = 0 \\ \: \: \: \\ \: \: \: \: \: \: \: \: \frac{ \sqrt{x} }{ \sqrt{16 + \sqrt{x} - 4} } \: \: \: x > 0$
is continuous at x=0

Answer 1

As its continuous

So lim x->0- (1 - cos 4x)/x^2

= 2 sin^2 2x/x^2

= 8 ( sin^2 2x/4x^2)

As sin^2 2x/4x^2 = 1

So it's 8

As at x= 1 its k

also continuous

So k= 8