54 lines
1.2 KiB
TeX
54 lines
1.2 KiB
TeX
\beginsection 19.2
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Prove that each of the following functions is uniformly continuous on the
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indicated set by directly verifying the $\epsilon$-$\delta$ property in
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Definition 19.1.
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\medskip
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To verify the $\epsilon$-$\delta$ property, start with the expression
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$|x-y|<\delta$ and then find a way to transform it into the expression
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$|f(x)-f(y)|<\epsilon$.
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\medskip
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(a) $f(x)=3x+11$ on $R$.
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\medskip
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We have
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$$|f(x)-f(y)|=|3x+11-3y-11|=3|x-y|$$
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Starting with
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$$|x-y|<\delta$$
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we can make the substitution
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$$|f(x)-f(y)|<3\delta$$
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Then for $\epsilon=3\delta$ we have
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$$|f(x)-f(y)|<\epsilon$$
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\medskip
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(b) $f(x)=x^2$ on $[0,3]$
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\medskip
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We have
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$$|f(x)-f(y)|=|x^2-y^2|=|x+y|\cdot|x-y|$$
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and
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$$|f(x)-f(y)|\le6|x-y|$$
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on the interval $[0,3]$.
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Starting with
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$$|x-y|<\delta$$
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we can make the substitution
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$$|f(x)-f(y)|<6\delta$$
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Then for $\epsilon=6\delta$ we have
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$$|f(x)-f(y)|<\epsilon$$
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\medskip
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(c) $f(x)=1/x$ on $[{1\over2},\infty)$
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\medskip
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We have
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$$|f(x)-f(y)|=\left|{1\over x}-{1\over y}\right|
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=\left|{y-x}\over xy\right|
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={|x-y|\over|xy|}$$
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and
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$$|f(x)-f(y)|\le4|x-y|$$
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on the interval $[{1\over2},\infty)$.
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Starting with
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$$|x-y|<\delta$$
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we can make the substitution
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$$|f(x)-f(y)|<4\delta$$
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Then for $\epsilon=4\delta$ we have
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$$|f(x)-f(y)|<\epsilon$$
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