Exercise 4.2.1

Question

\begin{enumerate}[label=(\alph*)] 
  \item Supply the details for how Corollary 4.2.4 part (ii) follows from the Sequential Criterion for Functional Limits in Theorem 4.2.3 and the Algebraic Limit Theorem for sequences proved in Chapter 2.
  \item Now, write another proof of Corollary 4.2.4 part (ii) directly from Definition 4.2.1 without using the sequential criterion in Theorem 4.2.3.
  \item Repeat (a) and (b) for Corollary 4.2.4 part (iii).
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item By the Sequential Criterion for Functional Limits, since $\lim_{x\to c} f(x) = L$, $\lim_{x\to c} g(x) = M$, all sequences $(x_n) \to c$ (where every $x_n \ne c$) have $f(x_n) \to L$ and $g(x_n) \to M$, which implies that $f(x_n) + g(x_n) \to L + M$ by the Algebraic Limit Theorem, which implies that $\lim_{x \to c} [f(x) + g(x)] = L + M$.
  \item Let $\epsilon > 0$, set $\delta_1$ such that $0<|x-c|<\delta_1$ implies $|f(x)-f(c)|<\epsilon/2$ and set $\delta_2$ such that $0<|x-c|<\delta_2$ implies $|g(x)-g(c)|<\epsilon/2$. Now let $\delta = \min\{\delta_1,\delta_2\}$ and use the triangle inequality to get
    $$
    |f(x)+g(x) - (f(c) + g(c))| \le |f(x)-f(c) | + |g(x)-g(c)| < \epsilon/2 + \epsilon/2 = \epsilon
    $$
    for all $0<|x-c|<\delta$.
  \item (a) is the same, if $f(x_n) \to L$ and $g(x_n) \to M$ then $(f(x_n)g(x_n)) \to LM$ by the sequential criterion for functional limits.

For (b) we add and subtract $f(c)g(x)$ then factor and use the triangle inequality (this is a common trick)
    $$
    \begin{aligned}
      |f(x)g(x) - f(c)g(c)| &= |g(x)(f(x)-f(c)) - f(c)(g(c)-g(x))| \
                            &\le |g(x)| |f(x)-f(c)| - |f(c)| |g(c)-g(x)|
    \end{aligned}
    $$
    Now we want a few things, (1) to bound $|g(x)|$ (2) to make $|f(x)-f(c)|$ small and (3) to make $|g(x)-g(c)|$ small. Whenever you want multiple things start thinking min/max!

In this case, set $\delta_1$ so $|g(x)-g(c)|<1$ giving the bound $|g(x)| < M+1$. Set $\delta_2$ so $|g(x)-g(c)|<\frac{\epsilon/2}{M+1}$ and set $\delta_3$ so $|f(x)-f(c)|< \frac{\epsilon/2}{f(c)}$. Finally set $\delta = \min\{\delta_1,\delta_2,\delta_3\}$ to get
    $$
    |f(x)g(x) - f(c)g(c)| < \epsilon/2 + \epsilon/2 = \epsilon
    $$
    as desired.
   \end{enumerate}

Exercise 4.2.1

Answers

Comments

Exercise 4.2.1

Answers

Comments

Add answer