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Secant slopes approximating g'(0) for g(x) = e^x

Consider the function g(x)=exg(x) = e^x. We want to estimate the instantaneous rate of change of gg at x=0x = 0 by computing average rates of change (secant slopes) over intervals that shrink toward 00.

The secant slope of g(x)=exg(x) = e^x through the points (0,g(0))(0, g(0)) and (h,g(h))(h, g(h)) is given by

g(h)−g(0)h=eh−1h.\frac{g(h) - g(0)}{h} = \frac{e^h - 1}{h}.

Compute this secant slope for h=0.1h = 0.1, h=0.01h = 0.01, and h=0.001h = 0.001 as scratch work (you do not need to report these values). Based on the trend you observe as h→0h \to 0, what value does this sequence of secant slopes appear to be approaching, and what does this value represent for g(x)=exg(x) = e^x at x=0x = 0?

Secant slopes approximating g'(0) for g(x) = e^x graph
Hint

Plug each value of hh into eh−1h\frac{e^h - 1}{h} using a calculator, and watch how the resulting number changes as hh gets smaller. Recall that this limiting value, if it exists, is by definition g′(0)g'(0).

Step-by-step solution

Substituting h=0.1h = 0.1, h=0.01h = 0.01, and h=0.001h = 0.001 into eh−1h\frac{e^h - 1}{h} gives values approximately 1.051711.05171, 1.005021.00502, and 1.00051.0005 respectively. Notice that as hh decreases toward 00, these secant slopes get closer and closer to a fixed number rather than diverging or oscillating.

The values are steadily approaching 11, suggesting that lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1. By the definition of the derivative, this limit is exactly g′(0)g'(0), the instantaneous rate of change of g(x)=exg(x) = e^x at x=0x = 0, which is also the slope of the tangent line to the curve at the point (0,1)(0,1).

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