## Calculus (3rd Edition)

$-\infty$
We have $$\lim _{x \rightarrow 3^+} \frac{x-4}{x^2-9}=\lim _{x \rightarrow 3^+} \frac{x-4}{(x+3)(x-3)}=- \infty.$$ We know that the limit from the right will be negative infinity because the numerator will be negative (since 3 is less than 4), while the denominator will be positive. The denominator will be arbitrarily close to zero, producing an infinite result.