Question

1 -cos 4x
lim
x →0 x2​

Answer 1

Answer:

$\lim_{x \to \00}\frac{(1-cos4x)}{x^2} \\ \lim_{x \to \00}\frac{(sin 2x^2)}{x^2} \\\\\ \lim_{x \to \00}\frac{(sin 2x^2)*4}{x^2*4} \\\\\since \lim_{x \to \00}\frac{(sinx)}{x} =1\\ \lim_{x \to \00}\frac{(sin 2x^2*4)}{x^2*4} \\\\\\=4$

Step-by-step explanation: