Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$ Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$
I'm I want to say that you cross multiply to get the same denominator, but I could be wrong.
Please Help!!

First, we find the common denominator:
$\require{cancel}$
$$\frac ab - \frac c{bd} = \frac{ad - c}{bd}$$
Applying the "general rule" in your case:
$$\begin{align} 
\frac 6x - \frac{42}{x^2 + 7x} & = \frac 6x - \frac{42}{x(x + 7)} \\ \\
& = \frac{6(x + 7) - 42}{x(x + 7)} \\ \\ & = \frac{6x}{x(x+7)} \\ \\ & = \frac{6\color{blue}{\cancel{x}}}{\color{blue}{\cancel{ x}}(x + 7)} \\ \\ & = \frac{6}{x + 7}
\end{align}$$
And so we have that $$\lim_{x\to 0} \frac 6x - \frac{42}{x^2 + 7x} \quad = \quad \lim_{x\to 0} \frac 6{x+7} \quad = \quad\frac 67$$
