Sketching a graph from limit conditions (asymptotes and jumps)

Text

Sketch the graph of a function $f(x)$ satisfying all the following conditions: $\lim_{x\to-\infty}f(x)=2^-,\qquad \lim_{x\to-1^-}f(x)=-\infty,\qquad \lim_{x\to-1^+}f(x)=+\infty,$ $\lim_{x\to0^{\pm}}f(x)=1,\ \ f(0)=3,\qquad \lim_{x\to2^-}f(x)=3^+,\qquad \lim_{x\to2^+}f(x)=-1^-,$ $\lim_{x\to+\infty}f(x)=+\infty.$

Solution

Let’s translate each condition into a feature of the graph.

• $\lim_{x\to-\infty}f(x)=2^-$: a left horizontal asymptote $y=2$, approached from below.
• $\lim_{x\to-1^-}f(x)=-\infty$ and $\lim_{x\to-1^+}f(x)=+\infty$: a vertical asymptote $x=-1$, with the left branch plunging to $-\infty$ and the right branch rising to $+\infty$.
• $\lim_{x\to0^{\pm}}f(x)=1$ but $f(0)=3$: at $x=0$ there is a removable discontinuity — an open dot at $(0,1)$ (the value the function approaches) and a filled dot at $(0,3)$ (the actual value of $f$).
• $\lim_{x\to2^-}f(x)=3^+$ and $\lim_{x\to2^+}f(x)=-1^-$: a jump (discontinuity of the first kind) at $x=2$, from the value $3$ (from above, arriving from the left) to the value $-1$ (from below, leaving toward the right).
• $\lim_{x\to+\infty}f(x)=+\infty$: on the right the graph diverges to $+\infty$.

$\boxed{\text{asymptotes } y=2,\ x=-1;\ \text{removable at }x=0;\ \text{jump at }x=2}$