Enter a T or an F in each answer space below to indicate whether the corresponding statement is true or false, the answers correct to receive credit.

$\underset{x1}{lim}\frac{x^{2}7x-18}{x^{2}6x-16}=\frac{\underset{x2}{lim}{x}^{2}7x-18}{\underset{x2}{lim}{x}^{2}6x-16}$

If $\underset{xf}{lim}|f(x)g(x)|$ exists, then the limit is $f(7)g(7)$

If $\underset{x6}{lim}f(x)=$ and $\underset{x6}{lim}g(x)=,$ then $\underset{x6}{lim}|f(x)-g(x)|=0$

If $\underset{x6}{lim}f(x)=0$ and $\underset{x6}{lim}g(x)=7,$ then $\underset{x6}{lim}|f\frac{x}{g}(x)|$ does not exist

If $f^{\text{'}}(7)$ exists, then then the limit $\underset{x!}{lim}f(x)$ is $f(7)$