## Thomas' Calculus 13th Edition

We know if $f=\dfrac{p}{q}$ is a rational function and $a$ is any real number, then $\lim\limits_{x \to a}f(x)=f(a)$ provided $q(a)$ is not zero. Hence since $\dfrac{2x+5}{11-x^3}$ is a rational function and $11-x^3 \neq 0$ for $x=2$, we have $\lim\limits_{x \to 2}\dfrac{2x+5}{11-x^3}=\dfrac{2(2)+5}{11-(2)^3}=\dfrac{4+5}{11-8}=\dfrac{9}{3}=3.$