Sketching a function from its limits (IV)

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Sketch the graph of a function $g(x)$ satisfying:

• (a) $\displaystyle\lim_{x \to -2^-} g(x) = -2^-$
• (b) $\displaystyle\lim_{x \to -2^+} g(x) = 2^-$
• (c) $\displaystyle\lim_{x \to -\infty} g(x) = -\infty$
• (d) $\displaystyle\lim_{x \to +\infty} g(x) = 0^-$

Solution

Reading:

• at $x=-2$ both one-sided limits are finite but different ($-2$ from the left, $2$ from the right): a jump at $x=-2$; moreover the superscripts indicate approach from below ($-2^-$ means tending to $-2$ staying just below, $2^-$ tending to $2$ staying just below);
• as $x\to-\infty$ the function diverges to $-\infty$;
• as $x\to+\infty$ it tends to $0$ from below: $y=0$ is a horizontal asymptote on the right, with the curve below the axis.

One possible graph: from the left it rises from $-\infty$ toward height $-2$ (approaching from below) just before $x=-2$; it jumps to just below $y=2$ right after, then descends toward $y=0$ staying negative. The solution is not unique.