Exercise 4.2.5

Question

Use Definition 4.2.1 to supply a proper proof for the following limit statements.
  \begin{enumerate}[label=(\alph*)] 
  \item $\lim _{x \rightarrow 2}(3 x+4)=10$
  \item $\lim _{x \rightarrow 0} x^{3}=0$
  \item $\lim _{x \rightarrow 2}\left(x^{2}+x-1\right)=5$.
  \item $\lim _{x \rightarrow 3} 1 / x=1 / 3$
   \end{enumerate}

Ulisse Mini · Accepted Answer

(Note that I use the largest $\delta$ choice that's easy to use)
  \begin{enumerate}[label=(\alph*)] 
  \item Since $|3x-6| = 3|x-2|$ setting $\delta = \epsilon/3$ gives $|3x-6|<\epsilon$ as desired.
  \item Since $|x^3| = |x|^3$ setting $\delta = \epsilon^{1/3}$ gives $|x|^3<\epsilon$ as desired.
  \item Since $|x^2+x-6| = |x-2||x+3| < \delta(5 + \delta)$ setting $\delta=\min\{1,\epsilon/6\}$ gives $\delta(5+\delta) < \delta(6) < \epsilon$ as desired.
    Another approach is to write $|x^2+x-6|$ in the ``$(x-2)^n$ basis''
    $$
    |x^2+x-6| = |(x-2)^2 + 5(x-2)| < \delta^2 + 5\delta = \delta(5 + \delta)
    $$
  \item We have $|1/x-1/3| = \frac{|3 - x|}{3|x|}$ setting $\delta = \min\{1,6\epsilon\}$ gives $1/3|x| < 1/6$ (because $|x| \in (2,4)$) and $|x-3|<6\epsilon$ meaning
    $$
    |1/x-1/3| = \frac{|3-x|}{3|x|} < \frac{|3-x|}{6} < \epsilon
    $$
    as desired.
   \end{enumerate}

Exercise 4.2.5

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Exercise 4.2.5

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