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\title{Math 165: Honors Calculus I\\
Assignment 16 Solutions}

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\problem{1} Show that $\lim_{x\to0^+}\sqrt{x} = 0$.

\medskip
\emph{Solution.}
Given $\epsilon>0$ we must find a $\delta>0$ (that may depend on
$\epsilon$) such that if $0 < x-0 < \delta$, then $|\sqrt{x} - 0| <
\epsilon$. Since $0<x<\delta$ implies that $0<\sqrt{x}<\sqrt{\delta}$
and we want to force $\sqrt{x}<\epsilon$, we should simply choose $\delta =
\epsilon^2$. Then $0 < x < \delta$ implies $0 < \sqrt{x} <
\sqrt{\epsilon^2} = \epsilon$.


\bigskip
\problem{2} Show that $\lim_{x\to0^+} x^{1/n} = 0$ for $n\in\N$.

\medskip
\emph{Solution.}
This is similar to Problem~1.  Given $\epsilon>0$ we must find a
$\delta>0$ such that if $0 < x-0 < \delta$, then $|x^{1/n} - 0| <
\epsilon$. Since $0<x<\delta$ implies that $0<x^{1/n}<\delta^{1/n}$ and
we want to force $x^{1/n}<\epsilon$, we should simply choose $\delta =
\epsilon^n$. Then $0 < x < \delta$ implies $|x|^{1/n} < (\epsilon^n)^{1/n}
= \epsilon$.

\bigskip
\problem{3} Show that $\lim_{x\to0^-} x^{1/n} = 0$ for $n$ an odd
positive integer.

\medskip
\emph{Solution.}
The proof is similar to the one in Problem~2, but we have to be careful
of negative numbers. For example, $n$ has to be odd to be able to define
$x^{1/n}$ for $x<0$. Given $\epsilon>0$ we must find a $\delta>0$ such
that if $-\delta < x < 0$, then $|x^{1/n} - 0| < \epsilon$. Since $-\delta
< x < 0$ implies $0 < |x| < \delta$ and $|x|^{1/n} < \delta^{1/n}$, and
since we want to force $|x|^{1/n}<\epsilon$, we should simply choose
$\delta = \epsilon^n$. Then $-\delta < x < 0$ implies $|x| <
\delta = \epsilon^n$ and hence $|x|^{1/n} < \epsilon$.

\bigskip
\problem{4} Let $f(x) = \frac{|x|}{x}$ for $x \ne 0$. Use one-sided limits
to show that $\lim_{x\to0} f(x)$ does not exist.

\medskip
\emph{Solution.}
If $x<0$, then $f(x) = \frac{|x|}{x} = -1$. Since we are allowed to
assume $x<0$ in the left-hand limit $\lim_{x\to0^-}f(x)$, it follows
that $\lim_{x\to0^-}f(x) = \lim_{x\to0^-} -1 = -1$. Similarly, if $x>0$,
then $f(x) = \frac{|x|}{x} = 1$. So $\lim_{x\to0^+}f(x)=\lim_{x\to0^+}1
= 1$. Since the left-hand limit does not equal the right hand limit,
the limit itself, $\lim_{x\to0}f(x)$, cannot exist.

\bigskip
\problem{5} Let $f(x) = x^2$ and let $p\in\R$.
  \begin{enumerate}[a)]
  \bigskip
  \item Show that if $|x - p| < 1$ then $|x| < |p| + 1$.

  \medskip\noindent
  \emph{Solution.} If $|x - p| < 1$, then $||x| - |p|| \le |x-p| < 1$.
  This implies $|p|-1 < |x| < |p|+1$.

  \bigskip
  \item Show that if $|x-p| < 1$ then $|x^2 - p^2| < (2|p|+1)|x - p|$.

  \medskip\noindent
  \emph{Solution.}  If $|x-p|<1$ then $|x+p| \le |x| + |p| < (|p| + 1) +
  |p| = 2|p|+1$ (by the triangle inequality and part a)). Therefore,
  if $|x-p|<1$, then $|x^2-p^2| = |x+p||x-p| < (2|p|+1)|x-p|$.

  \bigskip
  \item Show that $\lim_{x\to p}x^2 = p^2$.

  \medskip\noindent
  \emph{Solution.} We must show that given any $\epsilon>0$ there is a
  $\delta>0$ such that if $|x-p|<\delta$ then $|x^2-p^2|
  < \epsilon$.  From part b) we know that if $|x-p| <
  1$ then $|x^2-p^2| < (2|p|+1)|x-p|$.  We want to force
  $|x^2-p^2|<\epsilon$ and we can do this by making $(2|p|+1)|x-p|
  < \epsilon$, or $|x-p| < \epsilon/(2|p|+1)$.  Since we need
  two constraints to make this work, namely $|x-p|<1$ and $|x-p|<
  \epsilon/(2|p|+1)$, we choose $\delta$ to be the smaller: $\delta
  = \min(1,\epsilon/(2|p|+1))$. Therefore, if $|x-p|<\delta$, then
  $|x-p|<1$ \emph{and} $|x-p|<\epsilon/(2|p|+1)$. Together, these imply
  that $|x^2-p^2| < (2|p|+1)|x-p| < (2|p|+1)\cdot\epsilon/(2|p|+1)
  = \epsilon$.

  \end{enumerate}


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