11.22 Limit of function
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Use the \(\epsilon - \delta\) definition of limit to prove the following limits:
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\(\lim_{x \to 2} 3x + 4 = 10\)
\(\forall\epsilon>0,\exists\delta=\frac{\varepsilon}{3}>0\) such that \(0<|x-2|<\delta,x\in X\implies|3x+4-10|=\left|3x-6\right|<\epsilon\). 2. \(\lim_{x \to 0} x^3 = 0\)
\(\forall\epsilon>0,\exists\delta=\sqrt[3]{\varepsilon}>0\) such that \(0<|x|<\delta,x\in X\implies|x^3|<\epsilon\). 3. \(\lim_{x \to 2} x^2 + x - 1 = 5\)
\(\forall \epsilon > 0, \exists \delta > 0\) such that \(0<|x-2|<\delta,x\in X\).
We want \(|x^2+x-6|=\left|x+3\right|\left|x-2\right|<\epsilon\)
Thus we need to bound \(|x+3|\)
Let \(\delta=1\), then \(1<x<3\), then \(|x+3|<4\)
Then take \(\delta=\min\left\lbrace1,\frac{\varepsilon}{4}\right\rbrace\)
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Compute each limit or state that it does not exist:
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\(\lim_{x \to 2} \frac{|x - 2|}{x - 2}\)
Doesn't exist
Consider \(\lim_{x\to2^{+}}\frac{|x-2|}{x-2}=\lim_{x\to2^{+}}\frac{x-2}{x-2}=1\) and \(\lim_{x\to2^{-}}\frac{|x-2|}{x-2}=\lim_{x\to2^{-}}-\frac{x-2}{x-2}=-1\) 2. \(\lim_{x \to 1} \frac{|x - 2|}{x - 2}\)
\(\lim_{x\to1}\frac{|x-2|}{x-2}=\lim_{x\to1}-\frac{x-2}{x-2}=-1\) 3. \(\lim_{x \to 9} \frac{9 - x}{3 - \sqrt{x}}\)
\(\lim_{x\to9}\frac{9-x}{3-\sqrt{x}}=\lim_{x\to9}\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{3-\sqrt{x}}=\lim_{x\to9}3+\sqrt{x}=6\) 4. \(\lim_{x \to 1} \left( \frac{1}{x^2 - 1} - \frac{1}{x^4 - 1} \right)\)
\(\lim_{x\to1}\left(\frac{1}{x^2-1}-\frac{1}{x^4-1}\right)=\lim_{x\to1}\left(\frac{\left(x^2+1\right)-1}{x^4-1}\right)=\lim_{x\to1}\left(\frac{x^2}{x^4-1}\right)=\lim_{x\to1}\frac{x^2}{\left(x^2+1\left)\left(x^2-1\left)\right.\right.\right.\right.}=\frac{1}{\lim_{x\to1}\left(x^2+1\left)\lim_{x\to1}\left(x^2-1\left)\right.\right.\right.\right.}=\frac{1}{4\lim_{x\to1}\left(x-1\right)}\)
Doesn't exist
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Find the following limits:
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\(\lim_{x \to \infty} \frac{\sqrt{x^2 + 9}}{x + 3}\)
\(\lim_{x\to\infty}\frac{\sqrt{x^2+9}}{x+3}=\lim_{x\to\infty}\sqrt{\frac{x^2+9}{x^2+6x+9}}=\lim_{x\to\infty}\sqrt{\frac{1+\frac{9}{x^2}}{1+\frac{6}{x}+\frac{9}{x^2}}}=1\) 2. \(\lim_{x \to \infty} \left( \frac{3x^2 + 2x + 1}{x^2 - 3x + 2} \right)^4\)
\(\lim_{x\to\infty}\left(\frac{3x^2+2x+1}{x^2-3x+2}\right)^4=\left(\lim_{x\to\infty}\frac{3+\frac{2}{x}+\frac{1}{x^2}}{1-\frac{3}{x}+\frac{2}{x^2}}\right)^4=\left(\frac31\right)^4=81\) 3. \(\lim_{x \to \infty} \left( \frac{\sqrt{x^2 + 1} + 1}{\sqrt[3]{x^6 + 1}} \right)^2\)
\(\lim_{x\to\infty}\left(\frac{\sqrt{x^2+1}+1}{\sqrt[3]{x^6+1}}\right)^2=\lim_{x\to\infty}\left(\frac{\sqrt{x^2+1}}{\sqrt[3]{x^6+1}}+\frac{1}{\sqrt[3]{x^6+1}}\right)^2=\lim_{x\to\infty}\left(\frac{\sqrt{x^2+1}}{\sqrt[3]{x^6+1}}\right)^2=\lim_{x\to\infty}\frac{x^2+1}{\sqrt[3]{\left(x^6+1\right)^2}}=\lim_{x\to\infty}\frac{\frac{1}{x^2}}{\sqrt[3]{x^6+2+\frac{1}{x^6}^{}}}=\frac{1}{\lim_{x\to\infty}\sqrt[3]{x^6+2+\frac{1}{x^6}^{}}}=0\)
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Use sequential limit definition to prove that \(\lim_{x \to 0} \frac{1}{x}\) does not exist.
\(x_n=\frac1x\) \(0\)
\(y_n=\frac{1}{x+1}\) \(1\)