How to read one-sided limits and function values from a graph

Key concept: Graph questions test whether you can separate a limit from an actual function value. That difference matters a lot when there are open circles, jumps, or two filled points at the same x-value.

Step by step

From the graph information, interpret the points at $x=-2$ and $x=2$ carefully.

a. $\lim_{x\to -2^-} f(x)$

Approaching $-2$ from the left, the graph heads toward the point $( -2, 0 )$. So

$\lim_{x\to -2^-} f(x)=0$

b. $\lim_{x\to -2^+} f(x)$

Approaching $-2$ from the right, the graph also approaches the same visible value at $(-2,0)$. So

$\lim_{x\to -2^+} f(x)=0$

c. $\lim_{x\to 2^-} f(x)$

As $x$ approaches $2$ from the left, the graph tends toward the point $(2,2)$. Therefore

$\lim_{x\to 2^-} f(x)=2$

d. $f(2)$

The actual function value is given by the filled point at $x=2$, which is $(2,-1)$. So

$f(2)=-1$

Answers

• a. $0$
• b. $0$
• c. $2$
• d. $-1$

The key distinction is that limits describe what the graph approaches, while $f(2)$ is the value at the point itself.

Pitfall alert

A very common mistake is using the open-circle-looking point as the function value when the problem explicitly asks for $f(2)$. Another one is assuming the left and right limits must match. On graphs with jumps, they often do not. Always read each side separately.

Try different conditions

If the point $(2,2)$ were an open circle and $(2,-1)$ were the filled point, then the left-hand limit could still be $2$ while $f(2)$ would be $-1$. If both one-sided limits at a point are equal but the filled point sits somewhere else, that means the function is discontinuous there but the two-sided limit still exists.

Further reading

one-sided limit, function value from graph, jump discontinuity