Thank you Alan,
This is where I want it. As more bits and peices get added it would be good if it gets cleaned up into a neater thread but for now this is great.
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More limits
$$\small{\text{$\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{(x+4)(x-2)}{(x-4)(x-6)} =\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{x^2+2x-8}{x^2-10x+24}=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}}{ \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}}=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ 1 + \dfrac{2}{x}-\dfrac{8}{x^2}}{ 1 - \dfrac{10}{x}+\dfrac{24}{x^2}}= 9\cdot \dfrac{ 1 }{ 1 }=9 $}}\\\\$$
\small{\text{$\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{(x+4)(x-2)}{(x-4)(x-6)}
=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{x^2+2x-8}{x^2-10x+24}
=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}}{ \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}}
=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ 1 + \dfrac{2}{x}-\dfrac{8}{x^2}}{ 1 - \dfrac{10}{x}+\dfrac{24}{x^2}}= 9\cdot \dfrac{ 1 }{ 1 }=9 $}}\\\\
+118725 
