Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.1 The Idea of Limits - 2.1 Exercises - Page 59: 4

Answer

Let $\mathrm{A}(\mathrm{a}, \mathrm{f}(\mathrm{a}))$ be the given point on the graph. Take a point $\mathrm{B}$ on the graph, relatively close to the point $\mathrm{A}(\mathrm{a}, \mathrm{f}(\mathrm{a}))$, so that the x-coordinate equals a+h, for some value of h. The point $\mathrm{B}$ has coordinates $(\mathrm{a}+\mathrm{h}, \mathrm{f}(\mathrm{a}+\mathrm{h}))$. The line $\mathrm{A}\mathrm{B}$ is a secant of the graph and has slope $m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h}$ Now, observing m when B gets closer and closer to A on the graph, that is, when the number h approaches zero, we will be figuring out if $m_{A}=\displaystyle \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$ exists, and if it does, evaluating it. If the limit exists, we will have found the slope which the secants approach when we place B almost on top of A, that is, the slope of the line that has only one common point with the graph (the tangent at A.) In this way, we either find the slope of the tangent, or we conclude that the tangent at A can't be found (doesn't exist.)

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