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Estimating a limit from a table of values

Consider the function f(x)=x3−1x−1f(x) = \dfrac{x^3 - 1}{x - 1} Note that f(x)f(x) is undefined at x=1x = 1, since this would require dividing by zero. We want to understand how f(x)f(x) behaves for values of xx near 1, even though f(1)f(1) itself does not exist.

Using a calculator or by direct computation, evaluate f(x)=x3−1x−1f(x) = \dfrac{x^3-1}{x-1} at the values x=0.9, 0.99, 0.999x = 0.9,\ 0.99,\ 0.999 (approaching 1 from the left) and x=1.1, 1.01, 1.001x = 1.1,\ 1.01,\ 1.001 (approaching 1 from the right). Organize your results into a table showing xx and f(x)f(x) for these six values. Based on the trend in your table, what value does f(x)f(x) appear to be approaching as x→1x \to 1? That is, what is lim⁡x→1x3−1x−1?\lim_{x \to 1} \frac{x^3-1}{x-1}?

Hint

You don't need to simplify the formula algebraically — just plug in each xx value into f(x)f(x) directly and see what number the outputs are getting closer to from both sides.

Step-by-step solution

Computing f(x)f(x) at the left-approaching values gives f(0.9)=2.71f(0.9) = 2.71, f(0.99)=2.9701f(0.99) = 2.9701, and f(0.999)=2.997001f(0.999) = 2.997001. Computing at the right-approaching values gives f(1.1)=3.31f(1.1) = 3.31, f(1.01)=3.0301f(1.01) = 3.0301, and f(1.001)=3.003001f(1.001) = 3.003001. In both directions, the outputs are getting closer and closer to the number 3 as xx gets closer to 1.

Estimating a limit from a table of values — solution graph

Since the values of f(x)f(x) approach 3 from both the left and the right as x→1x \to 1, we conclude that lim⁡x→1x3−1x−1=3\lim_{x \to 1} \dfrac{x^3-1}{x-1} = 3, even though f(1)f(1) itself is undefined.

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