Is something wrong with the calculation below?
$$ \begin{align} \lim_{x \to \infty} \frac{4x^2}{x-2} &= \lim_{x \to \infty} \frac{4x}{1-2/x} \\ & = \frac{(\lim_{x \to \infty} 4x)}{(\lim_{x \to \infty} 1-2/x)} \\ & = \frac{(\lim_{x \to \infty} 4x)}{1} \\ & = \lim_{x \to \infty} 4x \\ & = \infty. \end{align} $$
I ask because if there isn't then the following would seem correct,
$$ \begin{align} \lim_{x \to \infty} \frac{4x^2}{x-2} - 4x &= \left( \lim_{x \to \infty} \frac{4x^2}{x-2} \right) - \left( \lim_{x \to \infty} 4x \right) \\ & = \left( \lim_{x \to \infty} 4x \right) - \left( \lim_{x \to \infty} 4x \right) \\ & = \lim_{x \to \infty} 4x - 4x \\ & = \lim_{x \to \infty} 0 \\ & = 0. \end{align} $$
But it is not, since
$$ \begin{align} \lim_{x \to \infty} \frac{4x^2}{x-2} - 4x &= \lim_{x \to \infty} \frac{4x^2 - 4x(x-2)}{x-2} \\ &= \lim_{x \to \infty} \frac{8x}{x-2} \\ &= \lim_{x \to \infty} \frac{8}{1-2/x} \\ &= 8. \\ \end{align} $$
In what way is this wrong? Where is the mistake?