University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 78: 58

Answer

To show that $\lim_{x\to2}h(x)\ne M$, choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-2|\lt\delta\Rightarrow|h(x)-M|\ge\epsilon$$
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Work Step by Step

To show that $\lim_{x\to2}h(x)\ne M$, choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-2|\lt\delta\Rightarrow|h(x)-M|\ge\epsilon$$ a) $\lim_{x\to2}h(x)\ne 4$ Let $\epsilon=1$ - For $x\gt2$, $h(x)=2$, meaning that $$|h(x)-4|=|2-4|=|-2|=2\gt1$$ That completes our proof. We can conclude now that $\lim_{x\to2}h(x)\ne4$ b) Prove that $\lim_{x\to2}h(x)\ne3$ Let $\epsilon=1/2$ Again, for $x\gt2$, $h(x)=2$, meaning that $$|h(x)-3|=|2-3|=|-1|=1\gt\frac{1}{2}$$ That completes our proof. We can conclude now that $\lim_{x\to2}h(x)\ne3$ c) Prove that $\lim_{x\to2}h(x)\ne2$ Let $\epsilon=0.2$ - For $x\lt2$, $h(x)=x^2$, meaning that $|h(x)-2|=|x^2-2|$ - Since $0\lt|x-2|\lt\delta$, that means $-\delta\lt x-2\lt\delta$, and $2-\delta\lt x\lt2+\delta$ Now we need to take a value of $x$ so that $x\lt2$ and $2-\delta\lt x\lt2+\delta$. I would take $x=2-(\delta/2)$, thus $x^2=(2-(\delta/2))^2=4-2\delta+(\delta^2/4)$ That means $$|h(x)-2|=|x^2-2|=|4-2\delta+(\delta^2/4)-2|=|2-2\delta+(\delta^2/4)|$$ The graph of the function $|h(x)-2|=|2-2\delta+(\delta^2/4)|$ is shown below. As we can see from the graph, as $\delta\ge1$, $|h(x)-2|\ge0.25\gt0.2$ That completes our proof. We can conclude now that $\lim_{x\to2}h(x)\ne2$
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