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HW 36

Ben Finch

E 36.1

Let $f:{\mathbb{R}}_{\ge 0}\to \mathbb{R}$ given by $f(x)=\sqrt{x}$. Show that $f$ is continuous at $a=9$.

Scratch work:
$|f(x)-L|<\epsilon$
$|\sqrt{x}-3|<\epsilon$

Then $x\in (8,10)$. $\sqrt{8}+3<\sqrt{x}+3<\sqrt{10}+3$. $\delta <\epsilon (3+\sqrt{8})$. $\delta =min(1,\epsilon (3+\sqrt{8}))$.

Proof.
Let $\epsilon >0$. Let $f:\mathbb{R}\ge 0\to \mathbb{R}$ given by $f(x)=\sqrt{x}$. Fix $\delta =min(1,\epsilon (3+\sqrt{8}))$. Let $9\in (4,14)$ which $\subseteq {\mathbb{R}}_{\ge 0}$. Assume that $0<|x-9|<\delta$. Then

Thus $\underset{x\to 9}{lim}f(x)=3=f(9)$. So the function is continuous at $x=9$. $◻$

E 36.2
Prove that if a limit exists, it is unique.

Proof.
Let $f:\mathbb{R}\to \mathbb{R}$ be a function. Let $a\in \mathbb{R}$. Let $L_1,L_2\in \mathbb{R}$. Assume that $li{m}_{x\to a}f(x)=L_1$ and $li{m}_{x\to a}f(x)=L_2$. Let $\epsilon >0$. There exists ${\delta }_1>0$ such that whenever $0<|x-a|<\delta$, $|f(x)-L_1|<\frac{\epsilon }{2}$. Similarly, there exists ${\delta }_2>0$ such that whenever $0<|x-a|<{\delta }_2$, then $|f(x)-L_2|<\frac{\epsilon }{2}$. Let $\delta =min({\delta }_1,{\delta }_2)$. Assume that $0<|x-9|<\delta$. Then

Thus, if a limit exists, it is unique. $◻$

E 36.3

a) We aim to show that for every $\epsilon >0$, there exists a $\delta >0$ such that for all $x$ in the domain of $f$, if $0<|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$.

Since $f(x)=1$ for $x$ in $[0,1]$ and $f(x)=0$ otherwise, for any $a\in \mathbb{R}-\{0,1\}$, we can choose $\delta$ small enough so that the interval $(a-\delta ,a+\delta )$ is entirely contained within either $(0,1)$ or $(-\mathrm{\infty },0)\cup (1,+\mathrm{\infty })$. This means that $f(x)$ will be constant (either 0 or 1) within this interval, and therefore, $|f(x)-f(a)|=0<\epsilon$ for all $x$ satisfying $0<|x-a|<\delta$. $◻$

b)
For continuity at $0$, we would need to find a $\delta$ for every $\epsilon >0$ such that $|f(x)-f(0)|<\epsilon$ whenever $0<|x-0|<\delta$. However, $f(0)=1$, and we can always find points $x$ close to 0 (specifically, any $x$ in $(-\delta ,0)$) where $f(x)=0$. No matter how small we make $\delta$, $|f(x)-f(0)|=|0-1|=1$ which is not less than $\epsilon$ if $\epsilon <1$. Thus, we cannot satisfy the condition for continuity at $0$, which means $f$ is not continuous at $0$. $◻$

E 36.4
We need to show that for every $\epsilon >0$, there exists a $\delta >0$ such that whenever $0<|x-a|<\delta$, implies $|f(x)g(x)-0|<\epsilon$.

Since $\underset{x\to a}{lim}f(x)=0$, for the given $\epsilon >0$, there exists a ${\delta }_1>0$ such that whenever $0<|x-a|<{\delta }_1$, it follows that $|f(x)-0|<\sqrt{\epsilon }$. Similarly, since $\underset{x\to a}{lim}g(x)=0$, there exists a ${\delta }_2>0$ such that whenever $0<|x-a|<{\delta }_2$, we have $|g(x)-0|<\sqrt{\epsilon }$.

We can take $\delta =min({\delta }_1,{\delta }_2)$. Then, for $0<|x-a|<\delta$, we have both $|f(x)|<\sqrt{\epsilon }$ and $|g(x)|<\sqrt{\epsilon }$. Therefore, for these values of $x$:

Hence, $|f(x)g(x)-LM|<\epsilon$ whenever $0<|x-a|<\delta$, so $\underset{x\to a}{lim}(f(x)g(x))=0$. $◻$

E 36.5
Proof.
Let the functions be as stated. Since $g(x)$ and $h(x)$ are polynomials, by theorem 36.14, they are both continuous at every real number. Then by theorem 36.12, $f(x)$ is also continous. $◻$