For \(f\colon D\to\RR\) with \(D\subseteq\RR\) and \(a\) a limit point of \(D\), \(\lim_{x\to a}f(x)=L\) means that for every \(\varepsilon>0\) there exists \(\delta>0\) such that \(\abs{f(x)-L}<\varepsilon\) for all \(x\in D\) with \(0<\abs{x-a}<\delta\). For a polynomial \(f\), write \(f(x)-f(a)=(x-a)g(x)\) with \(g\) a polynomial; if \(\abs{g}\le C\) on \(\abs{x-a}<1\), then \(\delta=\min(1,\varepsilon/C)\) works. The function \(f\) is continuous at \(a\in D\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that \(\abs{f(x)-f(a)}<\varepsilon\) for all \(x\in D\) with \(\abs{x-a}<\delta\). The function \(\abs{x}\) is continuous and not differentiable at \(0\), and the function equal to \(x^2\sin(1/x)\) for \(x\ne0\) and to \(0\) at \(0\) is differentiable on \(\RR\) with derivative discontinuous at \(0\).
$\lim_{x\to 2}(x+1/x)=5/2$ (Tag P-3OH6H)
Use the \(\varepsilon\)-\(\delta\) definition to prove that \[ \lim_{x\to2}\left(x+\frac1x\right)=\frac52. \]
Solution
Let \(\varepsilon>0\) and choose \[ \delta=\min\left\{1,\frac{2\varepsilon}{5}\right\}. \] If \(0<|x-2|<\delta\), then \(1<x<3\). Hence \(|x|>1\) and \(|2x-1|<5\). Therefore \[ \left|x+\frac1x-\frac52\right| =\frac{|2x-1|\,|x-2|}{2|x|} <\frac52|x-2| <\frac52\delta \leq\varepsilon. \]
$\lim_{x\to 2}\frac{1}{3+x}=\frac{1}{5}$ (Tag P-L3LHW)
Give an \(\varepsilon\)-\(\delta\) proof that \[ \lim_{x\to2}\frac1{3+x}=\frac15. \]
Solution
Let \(\varepsilon>0\) and choose \(\delta=\min\{1,20\varepsilon\}\). If \(|x-2|<\delta\), then \(1<x<3\), so \(|5(3+x)|>20\). Therefore \[ \left|\frac1{3+x}-\frac15\right| =\left|\frac{2-x}{5(3+x)}\right| <\frac1{20}|x-2| <\varepsilon. \]
$\lim_{x\to 1}\frac{x^2+1}{x}=2$ (Tag P-LF4NL)
Use the \(\varepsilon\)-\(\delta\) definition of the limit to prove that \[ \lim_{x\to1}\frac{x^2+1}{x}=2. \]
Solution
Let \(\varepsilon>0\) and choose \[ \delta=\min\left\{\frac12,\sqrt{\frac\varepsilon2}\right\}. \] If \(|x-1|<\delta\), then \(|x|>1/2\). Hence \[ \left|\frac{x^2+1}{x}-2\right| =\frac{|x-1|^2}{|x|} <2|x-1|^2 <2\delta^2 \leq\varepsilon. \]
The $\varepsilon$-$\delta$ definition of $\lim_{x\to a}f(x)=\ell$, and $\lim_{x\to 2}\frac{2x+1}{x^2+1}$ (Tag P-TBXVH)
Suppose \(f: \mathbb{R} \to \mathbb{R}\).
- Give the \(\delta\)-\(\varepsilon\) definition of \(\lim_{x \to a} f(x) = \ell\).
- Determine the limit and use the definition to prove your answer: \[\lim_{x \to 2} \frac{2x+1}{x^2+1} = ?\]
Solution
The statement \(\lim_{x\to a}f(x)=\ell\) means: for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ 0<|x-a|<\delta\implies |f(x)-\ell|<\varepsilon. \]
Here the candidate limit is \[ \frac{2(2)+1}{2^2+1}=1. \] Moreover \[ \left|\frac{2x+1}{x^2+1}-1\right| =\frac{|x(2-x)|}{x^2+1}. \] If \(|x-2|<1\), then \(|x|<3\), while \(x^2+1\ge1\). Thus \[ \left|\frac{2x+1}{x^2+1}-1\right|\le3|x-2|. \] Given \(\varepsilon>0\), choose \[ \delta=\min\{1,\varepsilon/3\}. \] Then \(0<|x-2|<\delta\) implies the displayed error is \(<\varepsilon\). Hence the limit is \(1\).
$\lim_{x\to 0}f(x)=0$ and $g$ bounded imply $f(x)g(x)\to 0$, with a counterexample if $g$ is unbounded (Tag P-TEVPO)
Suppose \(f\) and \(g\) are functions mapping \(\mathbb{R}\) into itself with \(\lim_{x\to 0} f(x) = 0\).
Prove from the \(\varepsilon\)–\(\delta\) definition that if \(g\) is bounded, then \(\lim_{x\to 0} f(x)g(x) = 0\) as well.
Give an example to show that the boundedness hypothesis cannot be omitted from part (a).
Solution
Suppose \(\exists M_g \suchthat \forall x,~ g(x) < M\). Then let \(\varepsilon > 0\) be arbitrarily chosen; we want to show that there exists a \(\delta\) such that \(\abs{x} \leq \delta \implies \abs{f(x)g(x)} \leq \varepsilon\). Since \(\lim_{x\to 0} f(x) = 0\), choose a \(\delta_f\) such that \(\abs{x} \leq \delta_f \implies \abs{f(x)} \leq \frac{\varepsilon}{M_g}\). So letting \(\delta = \delta_f\), we have \[ \abs{x} \leq \delta \implies \abs{f(x)g(x)} = \abs{f(x)} \abs{g(x)} \leq {\frac{\varepsilon}{M_g}}\abs{g(x)} \leq \frac{\varepsilon}{M_g}M_g = \varepsilon. \qed \] Let \(f(x) = x\) and \(g(x) = \frac{1}{x}\). Note that \(g(x)\) is unbounded in any neighborhood of 0, and \(f(x)g(x) = 1 \not\to 0\).
Examples - a false converse, continuous not differentiable, separate continuity, non-Cauchy bounded sequence, and a series converging exactly on $[0,2]$ (Tag P-FPWV6)
Provide examples of the following. [No justification is required.]
a true implication whose converse is false,
a function \(f: \mathbb{R} \to \mathbb{R}\) which is continuous at \(7\), but not differentiable there,
a function \(f: \mathbb{R}^2 \to \mathbb{R}\) which is continuous in each variable separately, but is not continuous at \((0,0)\),
a bounded sequence which is not Cauchy,
a (real) power series whose domain of convergence is the closed interval \([0,2]\).
Solution
<1>1. Part (a): A true implication whose converse is false: <2>1. Consider the statement for an integer \(n \in \mathbb{Z}\): \[ P(n) \implies Q(n), \quad \text{where } P(n) \text{ is } ``4 \mid n" \text{ and } Q(n) \text{ is } ``2 \mid n". \] If \(4 \mid n\), then \(n = 4k = 2(2k)\), so \(2 \mid n\), which is a true implication. <2>2. The converse \(Q(n) \implies P(n)\) is false, as witnessed by the counterexample \(n = 2\): \(2 \mid 2\) is true, but \(4 \mid 2\) is false.
<1>2. Part (b): Continuous at \(7\) but not differentiable at \(7\): <2>1. Define \(f: \mathbb{R} \to \mathbb{R}\) by: \[ f(x) = |x - 7|. \] \(f\) is continuous on \(\mathbb{R}\) as the composition of continuous functions. <2>2. Compute the one-sided derivative difference quotients at \(x = 7\): \[ \lim_{h \to 0^+} \frac{f(7+h) - f(7)}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = 1, \qquad \lim_{h \to 0^-} \frac{f(7+h) - f(7)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1. \] Because the left and right limits do not agree, \(f'(7)\) does not exist.
<1>3. Part (c): Separately continuous on \(\mathbb{R}^2\) but not continuous at \((0, 0)\): <2>1. Define \(f: \mathbb{R}^2 \to \mathbb{R}\) by: \[ f(x, y) = \begin{cases} \dfrac{xy}{x^2 + y^2} & (x, y) \neq (0, 0), \\ 0 & (x, y) = (0, 0). \end{cases} \] <2>2. For any fixed \(y_0\), \(x \mapsto f(x, y_0)\) is continuous on \(\mathbb{R}\) (at \(x = 0\), \(f(x, 0) = 0\) for all \(x\), so \(\lim_{x\to 0} f(x, 0) = 0 = f(0, 0)\)). Symmetrically, for any fixed \(x_0\), \(y \mapsto f(x_0, y)\) is continuous on \(\mathbb{R}\). Thus \(f\) is separately continuous in each variable. <2>3. Along the line \(y = x\), for \(x \neq 0\): \[ f(x, x) = \frac{x^2}{x^2 + x^2} = \frac{1}{2}. \] Thus \(\lim_{x \to 0} f(x, x) = \frac{1}{2} \neq 0 = f(0, 0)\), so \(f\) is not continuous at \((0, 0)\).
<1>4. Part (d): A bounded sequence that is not Cauchy: <2>1. Consider the sequence \((a_n)_{n=1}^\infty\) defined by \(a_n = (-1)^n\). Then \(|a_n| = 1 \le 1\) for all \(n\), so the sequence is bounded. <2>2. For any \(n \in \mathbb{N}\), \(|a_{n+1} - a_n| = |(-1)^{n+1} - (-1)^n| = 2\). Taking \(\varepsilon = 1\), there is no \(N\) such that \(|a_n - a_m| < 1\) for all \(n, m \ge N\). Thus \((a_n)\) is not Cauchy.
<1>5. Part (e): A real power series whose domain of convergence is \([0, 2]\): <2>1. Define the power series centered at \(c = 1\): \[ S(x) = \sum_{n=1}^\infty \frac{(x - 1)^n}{n^2}. \] <2>2. The radius of convergence \(R\) satisfies: \[ \frac{1}{R} = \lim_{n \to \infty} \left| \frac{1}{n^2} \right|^{1/n} = 1 \implies R = 1. \] Thus the series converges absolutely for \(|x - 1| < 1\) (i.e. \(x \in (0, 2)\)) and diverges for \(|x - 1| > 1\). <2>3. At the endpoints: - For \(x = 2\): \(\sum_{n=1}^\infty \frac{1^n}{n^2} = \sum_{n=1}^\infty \frac{1}{n^2}\), which converges (a \(p\)-series with \(p = 2 > 1\)). - For \(x = 0\): \(\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\), which converges absolutely by comparison to \(\sum \frac{1}{n^2}\). Therefore the exact interval of convergence is the closed interval \([0, 2]\).
<1>6. Conclusion: All five requested examples are defined and verified. Q.E.D.
Continuous solutions of $f(x)=5+\int_0^x 3f(t)\,dt$ (Tag P-UE5L6)
Suppose \(f\) is a continuous function satisfying the equation \(f(x) = 5 + \int_0^x 3f(t)\,dt\) for all real \(x\). Argue that \(f\) must be differentiable and then find all such function(s) explicitly.
Solution
<1>1. The function \(f\) is differentiable and satisfies \[ f'(x)=3f(x) \] for every \(x\in\mathbb R\).
Because \(f\) is continuous, the function \(3f\) is continuous. By the Fundamental Theorem of Calculus, \[ F(x)=\int_0^x 3f(t)\,dt \] is differentiable and \(F'(x)=3f(x)\). Since the given equation is \(f(x)=5+F(x)\), the function \(f\) is differentiable and \(f'(x)=3f(x)\).
<1>2. The initial value is \[ f(0)=5. \]
Substituting \(x=0\) into the integral equation gives \[ f(0)=5+\int_0^0 3f(t)\,dt=5. \]
<1>3. The only differentiable function satisfying <1>1 and <1>2 is \[ f(x)=5e^{3x}. \]
Define \(h(x)=e^{-3x}f(x)\). Then \[ h'(x)=e^{-3x}(f'(x)-3f(x))=0, \] so \(h\) is constant. Since \(h(0)=f(0)=5\), we have \(h(x)=5\) for all \(x\), hence \(f(x)=5e^{3x}\).
<1>4. This function indeed satisfies the original integral equation.
For \(f(x)=5e^{3x}\), \[ 5+\int_0^x 3f(t)\,dt =5+15\int_0^x e^{3t}\,dt =5+5(e^{3x}-1) =5e^{3x} =f(x). \]