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Estimating a limit numerically from a table of values

Consider the function f(x)=5x−5x−1f(x) = \dfrac{5^x - 5}{x - 1}. Note that f(x)f(x) is undefined at x=1x = 1, since plugging in x=1x=1 gives 00\frac{0}{0}. We want to understand the behavior of f(x)f(x) for xx values close to (but not equal to) 11.

Use a calculator or spreadsheet to compute f(x)=5x−5x−1f(x) = \dfrac{5^x-5}{x-1} for values of xx close to 11 on both sides, such as x=0.9,0.99,0.999x = 0.9, 0.99, 0.999 and x=1.1,1.01,1.001x = 1.1, 1.01, 1.001. 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, estimate lim⁡x→15x−5x−1.\lim_{x \to 1} \frac{5^x - 5}{x-1}.

Hint

You don't need to (and can't) plug in x=1x=1 directly, since that gives 00\frac{0}{0}. Instead, build a table with xx-values approaching 11 from the left and from the right, and watch what number the outputs get closer and closer to.

Step-by-step solution

Build a table of values approaching x=1x=1 from both sides. For example: f(0.9)≈7.83f(0.9)\approx 7.83, f(0.99)≈8.01f(0.99)\approx 8.01, f(0.999)≈8.045f(0.999)\approx 8.045, and from the right f(1.1)≈8.28f(1.1)\approx 8.28, f(1.01)≈8.07f(1.01)\approx 8.07, f(1.001)≈8.049f(1.001)\approx 8.049. Notice that as xx gets closer to 11 from either side, the outputs get closer to each other and settle near the same number.

Estimating a limit numerically from a table of values — solution graph

Both the left-hand and right-hand trends approach approximately 8.058.05, so the two-sided limit exists and equals this common value. (In fact this value equals 5ln⁡55\ln 5, though estimating it numerically to two or three decimal places is sufficient for this problem.)

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