Sketching a function from its limits (III)

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

• (a) $\displaystyle\lim_{x \to 2^-} g(x) = -3$
• (b) $\displaystyle\lim_{x \to 2^+} g(x) = 4$
• (c) $\displaystyle\lim_{x \to -\infty} g(x) = -\infty$
• (d) $\displaystyle\lim_{x \to +\infty} g(x) = 0^+$

Solution

Reading:

• at $x=2$ the left ($-3$) and right ($4$) limits are finite but different: there is a jump (first-kind discontinuity), no vertical asymptote;
• as $x\to-\infty$ the function diverges to $-\infty$;
• as $x\to+\infty$ it tends to $0$ from above, so $y=0$ is a horizontal asymptote on the right, with the curve above the axis.

One possible graph: from the left it rises from $-\infty$ up to height $-3$ just before $x=2$; it jumps to $4$ right after $x=2$ and then descends toward $y=0$ staying positive. The solution is not unique.