Find out for which value of k the function is continuous at x = 0

f(x) = (1-cos 4x) / x^2 when x ≠ 0

and f(x) = k(2+sin^2x) when x = 0

please tell me how to solve this

We need f(0) = lim(x→0) f(x)
<==> k(2 + sin^2(0)) = lim(x→0) (1 - cos(4x)) / x^2
<==> k = (1/2) * lim(x→0) (1 - cos(4x)) / x^2.

Using L'Hopital's Rule,
k = (1/2) * lim(x→0) 4 sin(4x) / (2x)
...= lim(x→0) 4 sin(4x)/(4x)
...= 4 * 1, via lim(t→0) sin(t)/t = 1 with t = 4x
...= 4.

I hope this helps!