dibantu ya kak plissss

[tex]\displaystyle \lim_{x\to0}\frac{1-\cos2x}{x^2}=\lim_{x\to0}\frac{2\sin^2x}{x^2}\\\lim_{x\to0}\frac{1-\cos2x}{x^2}=\lim_{x\to0}2\left(\frac{\sin x}{x}\right)^2\\\lim_{x\to0}\frac{1-\cos2x}{x^2}=2(1)^2\\\boxed{\boxed{\lim_{x\to0}\frac{1-\cos2x}{x^2}=2}}[/tex]

[tex]\displaystyle \lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=\lim_{x\to3}\frac{9-x}{x^2-9}-\frac{x+3}{x^2-9}\\\lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=\lim_{x\to3}\frac{9-x-x-3}{x^2-9}\\\lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=\lim_{x\to3}\frac{6-2x}{x^2-9}\\\lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=\lim_{x\to3}\frac{2(3-x)}{(x+3)(x-3)}\\\lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=\lim_{x\to3}\frac{-2}{x+3}\\\lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=\frac{-2}{3+3}[/tex]
[tex]\displaystyle \boxed{\boxed{\lim_{x\to3}\frac{9-x}{x^2-9}-\frac1{x-3}=-\frac{1}{3}}}[/tex]