Sketching a graph from limit conditions (second example)

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Sketch the graph of a function $g(x)$ satisfying: $\lim_{x\to-\infty}g(x)=+\infty,\qquad \lim_{x\to-3^-}g(x)=-1^-,\qquad \lim_{x\to-3^+}g(x)=-\infty,$ $\lim_{x\to1^-}g(x)=1^+,\qquad \lim_{x\to1^+}g(x)=1^-,\qquad \lim_{x\to+\infty}g(x)=2^-.$

Solution

• $\lim_{x\to-\infty}g(x)=+\infty$: on the left the graph diverges to $+\infty$.
• At $x=-3$: from the left the limit is the finite value $-1$ (approached from below, $-1^-$); from the right the branch drops to $-\infty$, so $x=-3$ is a vertical asymptote only for the right branch.
• At $x=1$: both one-sided limits equal $1$, but from opposite sides — $1^+$ from the left (from above), $1^-$ from the right (from below). The function crosses the value $1$ without forming a jump: the two branches “cross over” while staying close to $y=1$.
• $\lim_{x\to+\infty}g(x)=2^-$: a right horizontal asymptote $y=2$, approached from below.

$\boxed{\text{asymptotes } x=-3\ (\text{right branch}),\ y=2}$