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Estimating a finite limit from a table of values

Consider the function h(x)=2x−1xh(x) = \dfrac{2^x - 1}{x}, which is undefined at x=0x = 0 since it produces 00\frac{0}{0} there. We can still investigate what happens to h(x)h(x) as xx gets closer and closer to 00 from both sides.

Use a calculator or computer to evaluate h(x)=2x−1xh(x) = \dfrac{2^x - 1}{x} at values of xx close to 00, such as x=−0.1,−0.01,−0.001x = -0.1, -0.01, -0.001 and x=0.1,0.01,0.001x = 0.1, 0.01, 0.001. Based on the trend in your computed values, what does lim⁡x→02x−1x\displaystyle\lim_{x \to 0} \frac{2^x-1}{x} appear to equal?

Estimating a finite limit from a table of values graph
Hint

Plug values of xx approaching 00 from the left and from the right into h(x)h(x), and see what value the outputs seem to be getting closer to. Remember you never need to evaluate exactly at x=0x=0.

Step-by-step solution

Computing h(x)h(x) for xx near 00 from the right gives values like h(0.1)≈0.7177h(0.1) \approx 0.7177, h(0.01)≈0.6956h(0.01) \approx 0.6956, and h(0.001)≈0.6934h(0.001) \approx 0.6934. These values are steadily approaching a number close to 0.6930.693.

Computing h(x)h(x) for xx near 00 from the left gives h(−0.1)≈0.6698h(-0.1) \approx 0.6698, h(−0.01)≈0.6907h(-0.01) \approx 0.6907, and h(−0.001)≈0.6929h(-0.001) \approx 0.6929, which also approach approximately 0.6930.693. Since both sides approach the same value, we conclude the limit exists and lim⁡x→02x−1x≈0.693\displaystyle\lim_{x\to 0}\frac{2^x-1}{x} \approx 0.693 (which happens to equal ln⁡2\ln 2).

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