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Estimating a limit at a point discontinuity from a table of values

Consider the function f(x)={3x+1x≠210x=2f(x) = \begin{cases} 3x+1 & x \neq 2 \\ 10 & x = 2 \end{cases} Note that ff is defined differently right at x=2x=2 than it is everywhere nearby.

The table below shows values of f(x)f(x) for xx close to 22, approaching from both sides.

xx1.91.991.99922.0012.012.1
f(x)f(x)6.76.976.997107.0037.037.3

Using this table, determine lim⁡x→2f(x)\displaystyle\lim_{x \to 2} f(x). In your answer, also state whether this limit is equal to f(2)f(2).

Hint

Focus only on the values of f(x)f(x) for xx near 22 but not equal to 22 — ignore the middle column showing f(2)f(2) itself, since the limit only cares about nearby behavior.

Step-by-step solution

Looking at the table, as xx approaches 22 from the left (values 1.9,1.99,1.9991.9, 1.99, 1.999), f(x)f(x) approaches 77 from below (6.7,6.97,6.9976.7, 6.97, 6.997). As xx approaches 22 from the right (values 2.1,2.01,2.0012.1, 2.01, 2.001), f(x)f(x) also approaches 77, this time from above (7.3,7.03,7.0037.3, 7.03, 7.003). Since both one-sided trends converge to the same number, the two-sided limit exists and equals 77.

The actual value f(2)=10f(2) = 10 is irrelevant to the limit computation, since the limit depends only on values of ff near 22, not at 22. Therefore lim⁡x→2f(x)=7\lim_{x\to 2} f(x) = 7, while f(2)=10f(2) = 10, so the limit does not equal the function value at that point — this reflects a removable (point) discontinuity.

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