University Calculus: Early Transcendentals (3rd Edition)

$\lim_{x\to0^-}f(x)=-3$
If $f$ is an odd function of $x$, then $f(-x)=-f(x)$. First, since $\lim_{x\to0^+}f(x)$ exists means that $\lim_{x\to0^-}f(x)$ also exists. Because as $x$ approaches $0$ from the right and $f(x)$ approaches a value $X$, all coordinates $(x,f(x))$ must exist and approach a definite coordinate. So as $x$ approaches $0$ from the left, since $f(-x)=-f(x)$, all coordinates of $(-x,-f(x))$ must also exist and approach a definite coordinate. We know that $\lim_{x\to0^+}f(x)=3$. This means as $x$ approaches $0$ from the right $(x\gt0)$, the values of $f(x)$ gets closer and closer to $3$. So if $x$ approaches $0$ from the left $(x\lt0)$, as $-x$, $f(-x)=-f(x)$, the values of $f(x)$ now would approach the negative version of $f(x)$, which is $-3$. In other words, $\lim_{x\to0^-}f(x)=-3$ A graph of an odd function is shown below for clarification.