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!!
 A: You are correct that a common denominator is necessary:
\begin{align}
\frac 6 x - \frac {42} {x^2 + 7x} &= \frac 6 x - \frac {42} {x(x + 7)}\\
&= \frac{6(x + 7) - 42}{x(x + 7)} \\
&= \frac{6x}{x(x + 7)} \\
&= \frac{6}{x + 7}
\end{align}
for $x \ne 0$. Do you see how to compute the limit now?
A: $$\lim_{x \to 0} \frac1x \frac{6x+42-42}{x+7} = \lim_{x \to 0}\frac1x \frac{6x}{x+7} = \lim_{x \to 0}\frac{6}{x+7} = ?.$$
A: 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 $$\lim_{x\to 0} \frac 6x - \frac{42}{x^2 + 7x} \quad = \quad \lim_{x\to 0} \frac 6{x+7} \quad = \quad\frac 67$$
A: \begin{align}
{6 \over x} - {42 \over x^{2} + 7x}
&=
{6 \over x}\left(1 - {1 \over 1 + x/7}\right)
=
{6 \over x}
\left\lbrace
1
-
\left\lbrack 1 - {x \over 7} + {\rm O}\left(x^{2}\right)\right\rbrack
\right\rbrace
=
{6 \over x}\left\lbrack {x \over 7} + {\rm O}\left(x^{2}\right)\right\rbrack
\\[3mm]&=
{6 \over 7} + {\rm O}\left(x\right)
\color{#ff0000}{\large\to {6 \over 7}}
\end{align}
