Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises - Page 68: 76

Answer

Using graphical, numerical and analytic methods, it can be seen that: $\lim_{x\to16}\dfrac{4-\sqrt{x}}{x-16}=-\dfrac{1}{8}=-0.125$

Work Step by Step

$\lim_{x\to16}\dfrac{4-\sqrt{x}}{x-16}$ From the graph (shown in the answer section), $\lim_{x\to16}\dfrac{4-\sqrt{x}}{x-16}$ can be estimated as $-0.125$ To reinforce this estimation, elaborate a table (shown below), approaching $16$ frmo the left and from the right. It can be also seen that $\lim_{x\to16}\dfrac{4-\sqrt{x}}{x-16}=-0.125$ Let's confirm these estimations by evaluating the limit by analytic methods: $\lim_{x\to16}\dfrac{4-\sqrt{x}}{x-16}=...$ Factor the denominator as a difference of squares: $...=\lim_{x\to16}\dfrac{4-\sqrt{x}}{(\sqrt{x}-4)(\sqrt{x}+4)}=...$ Change the sign of the numerator and the sign of the fraction: $...=\lim_{x\to16}-\dfrac{\sqrt{x}-4}{(\sqrt{x}-4)(\sqrt{x}+4)}=...$ Take the minus sign out of the limit: $...=-\lim_{x\to16}\dfrac{\sqrt{x}-4}{(\sqrt{x}-4)(\sqrt{x}+4)}=...$ Simplify: $...=-\lim_{x\to16}\dfrac{1}{\sqrt{x}+4}=...$ Evaluate the limit applying direct substitution: $...=-\dfrac{1}{\sqrt{16}+4}=-\dfrac{1}{4+4}=-\dfrac{1}{8}$
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