Suppose $f(x)=\frac{x-3}{(x+3)(x+7)}$. Which of the following is not true?
$f$ has a vertical asymptote at $x=-3$
$f$ has a vertical asymptote at $x=-7$
$f$ has a removeable discontinuity at $x=3$.
$f$ has a horizontal intercept at $x=3$.
$f(x) \rightarrow 0$ as $x \rightarrow \pm \infty$

Step 1 ：First, we look at the function $f(x)=\frac{x-3}{(x+3)(x+7)}$.

Step 2 ：We can see that the function has vertical asymptotes at $x=-3$ and $x=-7$ because these are the values of $x$ that make the denominator equal to zero.

Step 3 ：There is no factor of $x-3$ in the denominator to cancel out the $x-3$ in the numerator, so there is no removable discontinuity at $x=3$.

Step 4 ：The function has a horizontal intercept at $x=3$ because this is the value of $x$ that makes the numerator equal to zero.

Step 5 ：As $x$ approaches $\pm \infty$, the function $f(x)$ approaches $0$ because the degree of the denominator is greater than the degree of the numerator.

Step 6 ：So, the statement that is not true is '$f$ has a removable discontinuity at $x=3$'.

Step 7 ：\(\boxed{'f$ has a removable discontinuity at $x=3$'}\) is the final answer.