Module 5

#Module 5: Limit Properties

\textbf{Example:} Let $f: \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = x^2$ for all $x \in \mathbb{R}$. Let $g: \mathbb{R} \to \mathbb{R}$ be the function defined by $g(x) = x^3$ for all $x \in \mathbb{R}$.

We can show that $\lim (f, 2, 4)$ and $\lim (g, 2, 8)$, i.e., that $$ \lim_{x \to 2} f(x) = 4 \quad \text{and} \quad \lim_{x \to 2} g(x) = 8 $$

(For a given $\varepsilon > 0$, $\delta_f = \min(1, \varepsilon/5)$ and $\delta_g = \min(1, \varepsilon/19)$ will work in the proof of the above limits).

\textbf{Question:} Can we use this information to determine $\lim_{x \to 2} (f(x) + g(x))$ ?

Can we use $\delta_f$ & $\delta_g$ to find the $\delta$ for this “sum” limit?

From graphing (& our knowledge concerning limits) we should get that:

$$ \lim_{x \to 2} (f(x) + g(x)) = \lim_{x \to 2} f(x) + \lim_{x \to 2} g(x) = 4 + 8 = 12. $$

\noindent Definition / Notation: Suppose we have $D \subseteq \mathbb{R}$ and $E \subseteq \mathbb{R}$, and functions $f: D \to E$ and $g: D \to E$.

\noindent We define a function $f+g$ as follows:

\noindent $(f+g): D \to E$

\noindent where: $x \mapsto f(x) + g(x)$ for all $x \in D$

\noindent or: $(f+g)(x) = f(x) + g(x)$

\noindent With this notation: We should have $\lim_{x \to 2} (f+g)(x) = 12$

\noindent Idea: $|(f+g)(x) - 12| < \epsilon \implies \text{This is what we will need.}$

\noindent $|(f+g)(x) - 12| = |f(x) + g(x) - 12| = |(f(x) - 4) + (g(x) - 8)|$

\noindent $\leq |f(x) - 4| + |g(x) - 8| \quad \text{(Triangle Inequality)}$

\noindent If we have $|f(x) - 4| < \epsilon/2$ and $|g(x) - 8| < \epsilon/2$, then this “$|f(x) - 4| + |g(x) - 8|$” will be less than $\frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

\noindent This is what we need!

We know that $ \lim_{x \to 2} f(x) = 4 $, so for any $ \varepsilon > 0$, there is a $ \delta_f > 0 $ such that if $ |x-2| < \delta_f$, then $ |f(x) - 4 | < \varepsilon$.

As $ \frac{\varepsilon}{2} > 0$, there is a corresponding $ \hat{\delta_f} > 0 $ such that if $ |x-2| < \hat{\delta_f} $, then $ |f(x) - 4 | < \frac{\varepsilon}{2} $.

Similarly, we know that $ \lim_{x \to 2} g(x) = 8 $, so for any $ \varepsilon > 0$, there is a $ \delta_g > 0$ such that if $ |x-2| < \delta_g$, then $ |g(x) - 8 | < \varepsilon $.

As $ \frac{\varepsilon}{2} > 0$, there is a corresponding $ \hat{\delta_g} > 0 $ such that if $ |x-2| < \hat{\delta_g} $, then $ |g(x) - 8 | < \frac{\varepsilon}{2}$.

If $ \delta = \min(\hat{\delta_f}, \hat{\delta_g})$, then if $ |x-2| < \delta$, then both $ |x-2| < \hat{\delta_f}$ and $ |x-2| < \hat{\delta_g}$.

From this, we have: $ |f(x) - 4| < \frac{\varepsilon}{2} $ and $ |g(x) - 8 | < \frac{\varepsilon}{2}$

Thus: $ |f(x) - 4 | + |g(x) - 8 | < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

\textbf{Theorem:} Let $D \subseteq \mathbb{R}$, $E \subseteq \mathbb{R}$. Also, let $f: D \to E$ and $g: D \to E$ be functions and let $a \in \mathbb{R}$ be an accumulation point of $D$. If $L \in \mathbb{R}$ and $M \in \mathbb{R}$ are such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then [ \lim_{x \to a} (f+g)(x) = L + M. ]

\textbf{Proof:} Let $D \subseteq \mathbb{R}$ and $E \subseteq \mathbb{R}$ be arbitrary. Also, let $f: D \to E$ and $g: D \to E$ be arbitrary functions. Let $a \in \mathbb{R}$ be an arbitrary accumulation point of $D$. Finally, let $L$ and $M$ be arbitrary real numbers such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$.

We need to show that $\lim_{x \to a} (f+g)(x) = L + M$, i.e., that [ (\forall \varepsilon > 0) (\exists \delta > 0) (\forall x \in D \setminus {a}) (|x-a| < \delta \implies |(f+g)(x) - (L+M)| < \varepsilon) ]

Let $\varepsilon > 0$ be arbitrary. As $\lim_{x \to a} f(x) = L$ and as $\frac{\varepsilon}{2} > 0$, there is some $\delta_f > 0$ such that whenever $x \in D \setminus {a}$ and $|x-a| < \delta_f$, then $|f(x) - L| < \frac{\varepsilon}{2}$.

Similarly, as $\lim_{x \to a} g(x) = M$ and $\frac{\varepsilon}{2} > 0$, there is some $\delta_g > 0$ such that whenever $x \in D \setminus {a}$ and $|x-a| < \delta_g$, then $|g(x) - M| < \frac{\varepsilon}{2}$.

Let $\delta = \min(\delta_f, \delta_g)$. As both $\delta_f > 0$ and $\delta_g > 0$, we have $\delta > 0$.

Now, let $x \in D \backslash {a}$ be arbitrary, and suppose $|x-a| < \delta$. As $\delta = \min(\delta_f, \delta_g)$, we have that $\delta \leq \delta_f$ and $\delta \leq \delta_g$, so that as $|x-a| < \delta$, we have both that $|x-a| < \delta_f$ and $|x-a| < \delta_g$.

As $|x-a| < \delta_f$, we have $|f(x) - L| < \epsilon/2$. Similarly, as $|x-a| < \delta_g$, we have $|g(x) - M| < \epsilon/2$.

Now: $$|f(x) - L| + |g(x) - M| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

By the triangle inequality: $$|(f(x) - L) + (g(x) - M)| \leq |f(x) - L| + |g(x) - M|.$$

Hence, $|(f+g)(x) - (L+M)| < \epsilon$. Therefore, if $|x-a| < \delta$, then $|(f+g)(x) - (L+M)| < \epsilon$. As $x \in D \backslash {a}$ was arbitrary, for all $x \in D \backslash {a}$, if $|x-a| < \delta$, then $|(f+g)(x) - (L+M)| < \epsilon$. As $\delta > 0$, there exists a $\delta > 0$ such that for all $x \in D \backslash {a}$, if $|x-a| < \delta$, then $|(f+g)(x) - (L+M)| < \epsilon$.

As $\epsilon > 0$ was arbitrary, for all $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x-a| < \delta$, then $|(f+g)(x) - (L+M)| < \epsilon$. Hence, we have that $\lim_{x \to a} (f+g)(x) = L + M$. As, $D, E, f, g, a, L$ and $M$ were arbitrary, for any sets $D \subseteq \mathbb{R}$ and $E \subseteq \mathbb{R}$, for any functions $f: D \to E$ and $g: D \to E$, any accumulation point $a$ of $D$ and any real numbers $L$ and $M$ such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, we have $\lim_{x \to a} (f+g)(x) = L + M$. $\blacksquare$

\begin{align*} &f(x) = \sqrt{x-5} \ &g(x) = (\sqrt{x-5})x \ &f: [5, \infty) \rightarrow \mathbb{R} \ &g: [5, \infty) \rightarrow \mathbb{R} \ &fg: [5, \infty) \rightarrow \mathbb{R} \ &(fg)(x) = f(x) \cdot g(x) \ &= (\sqrt{x-5})(\sqrt{x-5} x) \ &= (x-5)x \ &= x^2 - 5x \ &(fg)(0) = 0 \ & (fg)(0) \neq 0 \ & \frac{(fg)(0)}{} \ & DNE \ &(fg)(0) = f(0) \cdot g(0) \end{align*}

\begin{aligned} \lim_{x \to 0} xe^x &= (\lim_{x \to 0} x)(\lim_{x \to 0} e^x) \ &= 0 \cdot 1 \ &= 0 \end{aligned}

\begin{align*} & | f(x) \cdot g(x) - L g(x) + Lg(x) - LM | \ &= | g(x)(f(x) - L) + L(g(x) - M) | \ & |A+B| \leq |A| + |B| \ & \lim_{x \to a} f(x) = L \ & \forall \epsilon > 0 \quad \exists \delta > 0 \quad \forall x \in D \setminus {a } \quad (|x-a| < \delta \implies |f(x) - L| < \epsilon) \end{align*}

\begin{align*} 3 &= \frac{2}{3} + \frac{2}{3} + \frac{2}{3} \ &\ge | f(x) - f(x)| - | \ &\ge \frac{2}{3} | w - w(x) | \end{align*}

\begin{align*} & \text{goal} \ & ||g(x)-H|| \le \frac{\epsilon}{2} \ & |g(x) - H| \le \ & |L||g(x)-H| \le (L+1)|g(x)-H| \ & \frac{\epsilon}{3} \ge |g(x)-H| \ge \frac{\epsilon}{3} \ & |L| \le |L+1| \ & |Lg(x)-H| \ge (L+1)|g(x)-H| \ & |L| \le |L|+1 \ & |L||g(x)-H| \ge (L+1)|g(x)-H| \ & \frac{\epsilon}{3} \ge |g(x)-H| \end{align*}

|M| + 1 > M

$y = |g(x)|$

$\lim_{x \to a} g(x) = M$

$\lim_{x \to a} |g(x)| = |M|$

|M| - 1

a - $\delta$ a a + $\delta$

Then there is a $\delta$,

s/t If $|x - a| < \delta$

Then $|g(x)| - |M| | > 1$

$\exists \delta > 0 \forall x \in $

($ |x - a| < \delta$

I had trouble with some of the notation/symbols. I’ve made my best guess.

$S = S + \frac{(1 + 1 \times 2)}{(1 + 1)}S$

\noindent Definition / Notation: Suppose we have $D \subseteq \mathbb{R}$ and $E \subseteq \mathbb{R}$ and functions \ $F: D \to E$ and $g: D \to E$. \ We define a function $fg$ as follows: \ $(fg): D \to E$ \ where: $x \mapsto f(x) \cdot g(x)$ for all $x \in D$. \ or: $(fg)(x) = f(x) \cdot g(x)$ \ \qquad \qquad $\uparrow$ \qquad $\uparrow$ \ \qquad “New” Function \qquad “Old”/original/given functions \

\bigskip \noindent Our next goal will be to prove the following theorem: \ Let $D \subseteq \mathbb{R}$, $E \subseteq \mathbb{R}$ and let $f: D \to E$ and $g: D \to E$ be functions. \ Let $a \in \mathbb{R}$ be an accumulation point of $D$. Let $L \in \mathbb{R}$ and $M \in \mathbb{R}$. \ If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then $\lim_{x \to a} (fg)(x) = LM$. \

\bigskip \noindent As per usual, we will need to find $\delta$ such that if $|x-a| < \delta$, \ then: $| (fg)(x) - LM | < \epsilon$ \

\bigskip \noindent In Scratch Work we start here & work \ backwards to find $\delta = \delta(\epsilon)$

“Scratch Work”: \begin{align*} | (fg)(x) - LM | &= |f(x) \cdot g(x) - LM | \quad \xrightarrow[\text{}]{\text{Def. of } (fg)} \ &= |f(x) \cdot g(x) - L \cdot g(x) + L \cdot g(x) - LM | \ &= | g(x)(f(x) - L) + L(g(x) - M)| \end{align*}

Note: As $\lim_{x \to a} f(x) = L$, there is a $\delta_f > 0$ such that if $|x-a| < \delta_f$, then $|f(x)-L| < \epsilon$ (for any $\epsilon > 0$). As $\lim_{x \to a} g(x) = M$, there is a $\delta_g > 0$ such that if $|x-a| < \delta_g$, then $|g(x)-M| < \epsilon$ (for any $\epsilon > 0$).

\begin{align*} &\leq |g(x)| |f(x) - L | + |L | \cdot |g(x) - M| \quad \text{(Triangle Inequality)} \ &= |g(x)| |f(x) - L| + |L | \cdot |g(x) - M| \end{align*}

(1) We’ll try to make this $< \frac{\epsilon}{2}$:

(2) $|L| \cdot |g(x) - M | < (|L|+1) |g(x) - M| < \frac{\epsilon}{2}$

$|g(x) - M| < \frac{\frac{\epsilon}{2}}{(|L|+1)}$

Positive, so we get: $\delta_g = \delta_g \left( \frac{\epsilon}{2(|L|+1)} \right)$

\begin{align*} &x<y \ &3x<3y \ |x-a| < \delta &\implies |x-a| < \delta, \ &\implies |g(x)| - |M| < 1 \ &\implies -1 < |g(x)| - |M| < 1 \ &\implies |M|-1 < |g(x)| < |M| + 1 \ &\implies |g(x)| < |M| + 1 \ \text{Then: } |g(x)||f(x)| -L| &\leq (|M|+1)|f(x)-L| \tag{1}\ |x-a| < 5\delta &\implies |f(x) - L| < \frac{2(M+1)}{3} \ (|M|+1)|f(x) -L| &< \frac{2}{3} \tag{2} \ \text{Then } |g(x)||f(x)-L| &< \frac{2}{3} \end{align*}

|x-a| < \delta \longrightarrow |x-a| < \frac{5}{3}

\frac{|L|+1}{|(1+|g(x)-M|)} < |L|+1 \frac{2}{3} > \frac{|g(x)-M|}{\frac{(1+|L|+1)}{2/3}} \text{Then} \quad \frac{2}{3} > |115(x)-M|

\noindent ② We’ll try to get $|g(x)||f(x)-L| < \frac{\varepsilon}{2}$.

\noindent We’ll use a similar idea as in ①, but $|f(x)-L| < \frac{\varepsilon}{2(|g(x)|+1)}$ won’t work as this changes based on $x$ (so $\delta_\varepsilon$ would as well).

\noindent Instead: we will want $|g(x)| \leq M_g$, some positive constant.

\noindent If we get this, then: \begin{equation*} |g(x)||f(x)-L| \leq M_g|f(x)-L| < \frac{\varepsilon}{2} \end{equation*} \begin{equation*} |f(x)-L| < \frac{\varepsilon}{2M_g} \end{equation*} Positive, so we get $\delta_f = \delta_\varepsilon\left( \frac{\varepsilon}{2M_g} \right)$.

\vspace{0.3cm} \noindent \textbf{Notice:} As $1 > 0$, we can find a $\delta_1 > 0$ such that if $|x-a|<\delta_1$, then $||g(x)|-|M||<1$.

\noindent In particular, if $|x-a|<\delta_1$, then $|g(x)| < |M|+1 = M_g$

\noindent (We will need to prove that if $\lim_{x\to a} g(x) = M$, then $\lim_{x\to a} |g(x)| = |M|$).

\noindent (For now, we will assume this fact.)

“Idea” so far: We will three upper limits for $\delta$: [ \delta \leq \delta_1: \text{ Whenever } |x-a| < \delta_1, \text{ then } | |g(x)| - |M| | < 1 ] This will give $|g(x)| < |M| + 1$ [ \delta \leq \delta_f= \delta_f(\frac{\varepsilon}{2(|M|+1)}): \text{ When ever } |x-a| < \delta_f, \text{ then } |f(x) - L| < \frac{\varepsilon}{2(|M|+1)} ]

This will give $|g(x)| |f(x) - L| < \frac{\varepsilon}{2} $ [ \delta \leq \delta_g = \delta_g(\frac{\varepsilon}{2(|L|+1)}): \text{ Whenever } |x-a| < \delta_g, \text{ then } |g(x) - M| < \frac{\varepsilon}{2(|L|+1)} ]

This will give $|L||g(x) - M| < \frac{\varepsilon}{2} $

\underline{Putting this together}: We let $\delta = \min(\delta_1, \delta_f, \delta_g)$. For $|x-a| < \delta$:

As $|x-a| < \delta_1$, we have $| |g(x)| - |M| | < 1$, so $-1 < |g(x)| - |M| < 1$, and $|M| - 1 < |g(x)| < |M| + 1$, so $|g(x)| < |M| + 1$.

As $|x-a| < \delta_f$, we have $|f(x) - L| < \frac{\varepsilon}{2(|M|+1)}$, so $(|M|+1) |f(x) - L| < \frac{\varepsilon}{2}$.

As $|g(x)| < |M| + 1$, $|g(x)| |f(x) - L| < (|M|+1) |f(x) - L|$. Hence, $|g(x)||f(x) - L| < \frac{\varepsilon}{2}$

As $|x-a| < \delta_g$, we have $|g(x) - M| < \frac{\varepsilon}{2(|L|+1)}$, so $(|L|+1) |g(x) - M| < \frac{\varepsilon}{2}$.

As $|L| < |L|+1$, $|L||g(x) - M| < (|L|+1)|g(x) - M|$. Hence, $|L||g(x) - M| < \frac{\varepsilon}{2}$.

If $\lim_{x \to a} g(x) = M$, then $\lim_{x \to a} |g(x)| = |M|$.

\text{If } |x-a| < \delta, \text{ then } | l(x) - l | < 3

\noindent Putting this together: $|g(x)f(x) - LM| = |g(x)f(x) - Lg(x) + Lg(x) - LM| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$

\noindent \textbf{For the proof:} Set-up/introduce D, E, f and g. Introduce a, L, M. Assume $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$. Let $\epsilon > 0$ be arbitrary.

\noindent \begin{itemize} \item Use these assumed limits to define $\delta_f$ (for $\frac{\epsilon}{2(|M| + 1)}$), $\delta_g$ (for $\frac{\epsilon}{2(|L| + 1)}$), and $\delta_1$ (for 1). \item Set $\delta = \min(\delta_f, \delta_g, \delta_1)$ and explain why $\delta > 0$. \item Let $x \in D \setminus {a}$ be arbitrary and suppose $|x - a| < \delta$. \item Show $|(fg)(x) - LM| < \epsilon$. (Using the above arguments) \item Conclude the proof (building back up to quantified statement). \end{itemize}

\noindent Proof: Let $D \subseteq \mathbb{R}$, $E \subseteq \mathbb{R}$. Let $g : D \to E$ be a function and let $a \in \mathbb{R}$ be an accumulation point of $D$. Also, let $M \in \mathbb{R}$ and suppose $\lim_{x \to a} g(x) = M$.

\noindent We will prove that $\lim_{x \to a} |g(x)| = |M|$.

\noindent Let $\varepsilon > 0$ be arbitrary. As $\lim_{x \to a} g(x) = M$, there is a $\hat{\delta} > 0$ such that for any $x \in D \setminus {a}$, if $|x - a| < \hat{\delta}$, then $|g(x) - M| < \varepsilon$.

\noindent We will consider three cases.

\noindent Case 1: Suppose $M = 0$. Then, let $\delta = \hat{\delta}$. As $\hat{\delta} > 0$, $\delta > 0$. Let $x \in D \setminus {a}$ be arbitrary, and suppose $|x - a| < \delta$. Then, we have that $|x - a| < \hat{\delta}$ so that $|g(x) - M| < \varepsilon$. In this case: $|g(x) - 0| < \varepsilon$, so $|g(x)| < \varepsilon$, and $||g(x)|| < \varepsilon$. From this, $||g(x)| - 0| < \varepsilon$, and $||g(x)| - |0|| < \varepsilon$, so that $| |g(x)| - |M| | < \varepsilon$. Hence, if $|x - a| < \delta$, then $||g(x)| - |M|| < \varepsilon$.

\noindent As $x \in D \setminus {a}$ was arbitrary, for any $x \in D \setminus {a}$, if $|x - a| < \delta$, then $||g(x)| - |M|| < \varepsilon$. As $\delta > 0$, there exists $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x - a| < \delta$, then $||g(x)| - |M|| < \varepsilon$.

Case 2: Suppose $M > 0$. Note that $\frac{M}{2} > 0$ so that as $\lim_{x\to a} g(x) = M$, there is a $\hat{\delta} > 0$ such that for any $x \in D \setminus {a}$, if $|x-a| < \hat{\delta}$, then $|g(x) - M| < \frac{M}{2}$. Let $\delta = \min{\hat{\delta}, \tilde{\delta}}$. As $\tilde{\delta} > 0$ and as $\hat{\delta} > 0$, we have $\delta > 0$. Now, let $x \in D \setminus {a}$ be arbitrary, and suppose $|x-a| < \delta$. Then $|x-a| < \hat{\delta}$ so that:

$$|g(x) - M| < \frac{M}{2}$$ $$-\frac{M}{2} < g(x) - M < \frac{M}{2}$$ $$\frac{M}{2} < g(x) < \frac{3M}{2}.$$

As $\frac{M}{2} < g(x)$ and $\frac{M}{2} > 0$, we have that $g(x) > 0$, so that $|g(x)| = g(x)$.
As $M > 0$, we have that $|M| = M$.

Now, as $|x-a| < \tilde{\delta}$, we have that $|x-a| < \tilde{\delta}$ so that $||g(x)| - |M|| < \epsilon$.

Hence, $||g(x)| - |M|| < \epsilon$. Thus, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$. As $x \in D \setminus {a}$ was arbitrary, for any $x \in D \setminus {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$. As $\tilde{\delta} > 0$, there exists $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$.

\noindent Proof: Let $D \subseteq \mathbb{R}, E \subseteq \mathbb{R}$. Let $g: D \rightarrow E$ be a function and let $a \in \mathbb{R}$ be an accumulation point of $D$. Also, let $M \in \mathbb{R}$ and suppose $\lim_{x \rightarrow a} g(x) = M$.

\noindent We will prove that $\lim_{x \rightarrow a} |g(x)| = |M|$.

\noindent Let $\epsilon > 0$ be arbitrary. As $\lim_{x \rightarrow a} g(x) = M$, there is a $\hat{\delta} > 0$ such that for any $x \in D \setminus {a}$, if $|x - a| < \hat{\delta}$, then $|g(x) - M| < \epsilon$.

\noindent We will consider three cases.

\noindent Case 1: Suppose $M = 0$. Then, let $\delta = \hat{\delta}$. As $\hat{\delta} > 0$, $\delta > 0$. Let $x \in D \setminus {a}$ be arbitrary, and suppose $|x - a| < \delta$. Then, we have that $|x - a| < \hat{\delta}$ so that $|g(x) - M| < \epsilon$. In this case: $|g(x) - 0| < \epsilon$, so $|g(x)| < \epsilon$, and $||g(x)|| < \epsilon$. From this, $||g(x)| - |0|| < \epsilon$, and $||g(x)| - 0|| < \epsilon$, so that $||g(x)| - |M|| < \epsilon$. Hence, if $|x - a| < \delta$, then $||g(x)| - |M|| < \epsilon$.

\noindent As $x \in D \setminus {a}$ was arbitrary, for any $x \in D \setminus {a}$, if $|x - a| < \delta$, then $||g(x)| - |M|| < \epsilon$. As $\delta > 0$, there exists $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x - a| < \delta$, then $||g(x)| - |M|| < \epsilon$.

\begin{align*} &y = g(x)\ &\frac{3}{2}M\ &M\ &\frac{1}{2}M\ &a - \delta\ &a\ &a+\delta \end{align*}

\begin{tikzpicture} % Define nodes for labels \node (M) at (1, 3) {$M$}; \node (halfM) at (3, 3) {$\frac{|M|}{2}$}; \node (absM) at (5, 3) {$|M|$};

\node (a) at (3,1) {$a$}; \node (function) at (3, -2) {$f=g(x)$};

% Define a dummy node for graph (using tikzplot package) \end{tikzpicture}

\noindent Case 2: Suppose $M > 0$. Note that $\frac{M}{2} > 0$ so that as $\lim_{x \to a} g(x) = M$, there is a $\hat{\delta} > 0$ such that for any $x \in D \setminus {a}$, if $|x-a| < \hat{\delta}$, then $|g(x) - M| < \frac{M}{2}$. Let $\delta = \min(\hat{\delta}, \tilde{\delta})$. As $\tilde{\delta} > 0$ and as $\hat{\delta} > 0$, we have $\delta > 0$. Now, let $x \in D \setminus {a}$ be arbitrary, and suppose $|x-a| < \delta$. Then $|x-a| < \hat{\delta}$ so that:

\begin{align*} |g(x)| - M < \frac{M}{2} \ -\frac{M}{2} < g(x) - M < \frac{M}{2} \ \frac{M}{2} < g(x) < \frac{3M}{2}. \end{align*}

\noindent As $\frac{M}{2} < g(x)$ and $\frac{M}{2} > 0$, we have that $g(x) > 0$, so that $|g(x)| = g(x)$. As $M > 0$, we have that $|M| = M$.

\noindent Now, as $|x-a| < \tilde{\delta}$, we have that $|x-a| < \tilde{\delta}$ so that $|g(x)| - |M| < \epsilon$.

\noindent Hence, $||g(x)| - |M|| < \epsilon$. Thus, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$. As $x \in D \setminus {a}$ was arbitrary, for any $x \in D \setminus {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$. As $\delta > 0$, there exists $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$.

\textbf{Case 3:} Suppose that $M < 0$. Note that $\frac{|M|}{2} > 0$ so that as $\lim_{x \to a} g(x) = M$, there is a $\hat{\delta} > 0$ such that for any $x \in D \backslash {a}$, if $|x - a| < \hat{\delta}$, then $|g(x) - M| < \frac{|M|}{2}$. Let $\delta = \min(\hat{\delta}, \tilde{\delta})$. As $\hat{\delta} > 0$ and $\tilde{\delta} > 0$, we have $\delta > 0$. Now, let $x \in D \backslash {a}$ be arbitrary, and suppose $|x-a| < \delta$. Then, we have that $|x-a| < \hat{\delta}$ so that: $$ |g(x) - M| < \frac{|M|}{2} $$ $$ -\frac{|M|}{2} < g(x) - M < \frac{|M|}{2} $$ $$ M - \frac{|M|}{2} < g(x) < M + \frac{|M|}{2} $$

As $M < 0$, $|M| = -M$, so that: $$g(x) < M + \frac{|M|}{2} = M - \frac{M}{2} = \frac{M}{2}$$ Thus, $g(x) < \frac{M}{2}$. As $M < 0$, $\frac{M}{2} < 0$ so that $g(x) < 0$. Thus, $|g(x)| = -g(x)$.

Now, as $|x-a| < \delta$, we have $|x-a| < \tilde{\delta}$ so that $|g(x) - M| < \varepsilon$. Notice: \begin{align*} | |g(x)| - |M| | &= | 1 \cdot |g(x)| - |M| | = | -1 \cdot (|g(x)| - |M|)| \ &= |- g(x) - (-M)| = | -g(x) + M | = | - (g(x) - M)| \end{align*}

Hence, $||g(x)| - |M|| < \varepsilon$. Thus, if $|x-a| < \delta$, then $||g(x)| - |M|| < \varepsilon$. As $x \in D \backslash {a}$ was arbitrary for any $x \in D \backslash {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \varepsilon$. As $\delta > 0$, there exists $\delta > 0$ such that for all $x \in D \backslash {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \varepsilon$.

These cases are exhaustive. Hence, there exists a $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$. As $\epsilon > 0$ was arbitrary, for all $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in D \setminus {a}$, if $|x-a| < \delta$, then $||g(x)| - |M|| < \epsilon$. Therefore, $\lim_{x \to a} |g(x)| = |M|$.

As $D, E, g, a$ and $M$ were arbitrary, for any $D \subseteq \mathbb{R}, E \subseteq \mathbb{R}$ and any function $g: D \to E$ and any accumulation point $a \in \mathbb{R}$ of $D$ and any $M \in \mathbb{R}$, if $\lim_{x \to a} g(x) = M$, then $\lim_{x \to a} |g(x)| = |M|$.

\noindent \textbf{The Triangle Inequality:} For all real numbers $x$ and $y$, $|x+y| \leq |x| + |y|$.

\noindent \textbf{Proof:} Let $x$ and $y$ be arbitrary real numbers. We will consider four cases.

\noindent \textbf{Case 1:} Suppose $x < 0$ and $y < 0$. Then, $|x| = -x$ and $|y| = -y$. Also, $x+y < 0$, so $|x+y| = -(x+y)$. Now:

\noindent $|x+y| = -(x+y) = -x - y = (-x) + (-y) = |x| + |y|$.

\noindent Hence, $|x+y| = |x| + |y|$, so that $|x+y| \leq |x| + |y|$.

\noindent \textbf{Case 2:} Suppose $x < 0$ and $y \geq 0$. Then, $|x| = -x$ and $|y| = y$. We will consider two subcases.

\noindent \textbf{Subcase 2.1:} Suppose $x+y \geq 0$. Then, $|x+y| = x+y$. Now:

\noindent $|x+y| = x+y \leq 2|x| + x + y = -2x + x + y = -x+y = |x| + |y|$.

\noindent \textbf{Subcase 2.2:} Suppose $x+y < 0$. Then $|x+y| = -(x+y)$. Now:

\noindent $|x+y| = -(x+y) = -x -y \leq -x-y + 2|y| = -x-y+2y=-x+y = |x|+|y|$.

\noindent These subcases are exhaustive, so we have that $|x+y| \leq |x| + |y|$.

\noindent \textbf{Case 3:} Suppose $x \geq 0$ and $y < 0$. Notice that $|x+y| = |y+x|$, so by Case 2, we have $|y+x| \leq |y| + |x| = |x| + |y|$. Hence, $|x+y| \leq |x| + |y|$.

a = b = c < d = e = f \ a < f \ a < f

Case 4: Suppose $x \geq 0$ and $y \geq 0$. Then $|x| = x$ and $|y| = y$. Also $x+y \geq 0$ so that $|x+y| = x+y$. Now: $$|x+y| = x+y = |x| + |y|.$$ Thus, $|x+y| = |x|+|y|$, so that $|x+y| \leq |x|+|y|$. These cases are exhaustive, hence $|x+y| \leq |x|+|y|$. As $x$ and $y$ were arbitrary, for all real numbers $x$ and $y$, $|x+y| \leq |x|+|y|$. \newline \textbf{The Reverse Triangle Inequality:} For all real numbers $x$ and $y$, we have that: $$||x| - |y|| \leq |x-y|.$$ \textbf{Proof:} Let $x$ and $y$ be arbitrary real numbers. We will consider four cases.

\textit{Case 1:} Suppose $x < 0$ and $y < 0$. Then, $|x| = -x$ and $|y| = -y$. Now: $$||x| - |y|| = |-x+y| = |-(x-y)| = ||-(x-y)|| = |x-y|.$$ Hence, we have $||x| - |y|| = |x-y|$, so that $||x| - |y|| \leq |x-y|$.

\textit{Case 2:} Suppose $x < 0$ and $y > 0$. Then, $|x| = -x$ and $|y| = y$. Now: $$||x| - |y|| = |-x-y| = |-(x+y)| = |x+y|.$$ By the triangle inequality, $|x+y| \leq |x|+|y|$, so we have: $$||x| - |y|| = |x+y| \leq |x|+|y| = |-x+y| = |-(x-y)| = |x-y|.$$

Hence, we have $||x|-|y|| \leq |x-y|$.

Case 3: Suppose $x \geq 0$ and $y < 0$. Then, note that by case 2, $| |y| - |x| | \leq |y-x|$. Also, we have that: $| |x| - |y| | = | -(|x| - |y|) | = | |y| - |x| |$, and also that: $|x-y| = |-(x-y)| = |y-x|$. Hence, we have that $||x|-|y|| \leq |x-y|$.

Case 4: Suppose $x \geq 0$ and $y \geq 0$. Then $|x| = x$ and $|y| = y$. Now: $||x|-|y|| = |x-y|$. Thus, $||x|-|y|| \leq |x-y|$.

These cases are exhaustive, hence: $||x|-|y|| \leq |x-y|$. As $x$ and $y$ were arbitrary, for all real numbers $x$ and $y$, $||x|-|y|| \leq |x-y|$.

\lim_{x\to a} g(x) = M \implies \lim_{x\to a} |g(x)| = |M|

\forall \epsilon>0 ; \exists \delta>0 ; \forall x ; (|x-a|<\delta \implies |g(x) - M| < \epsilon )

\text{By the Reverse Triangle Inequality:}

\text{So } ||g(x)| - |M|| < \epsilon \text{ whenever } |x-a| < \delta

\begin{align*} &\lim_{x \to a} (f+g)(x) = \lim_{x \to a}f(x) + \lim_{x \to a}g(x) \ &\lim_{x \to a} (f g)(x) = \left( \lim_{x \to a}f(x) \right) \left( \lim_{x \to a}g(x) \right) \ &\lim_{x \to a} \left( \frac{f}{g} \right) (x) = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \end{align*}

(\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in Dom(\frac{f}{g}) \setminus {a}) (|x-a| < \delta \implies |\frac{f(x)}{g(x)} - \frac{L}{M}| < \epsilon)

\lim_{x \to a} f(x) = L

\lim_{x \to a} g(x) = M

(\exists > |M - g(x)| \Leftarrow \delta > |a-x|) \land x \in A \land \exists A \quad (\exists > |L - f(x)| \Leftarrow \delta > |a-x|) \land x \in A \land \exists A

\begin{aligned} \left| \frac{f(x)}{g(x)} - \frac{L}{M} \right| &= \left| \frac{M \cdot f(x) - L \cdot g(x)}{M \cdot g(x)} \right| \ &= \left| \frac{M \cdot f(x) - ML + ML - L \cdot g(x)}{M \cdot g(x)} \right| \ &= \left| \frac{M(f(x) - L) - L(g(x) - M)}{M \cdot g(x)} \right| \end{aligned}

\begin{align*} \left| \frac{M(f(x)) - L}{M(g(x))} - L \frac{(g(x) - M)}{M(g(x))} \right|
&= \left| \frac{M(f(x)) - L}{M(g(x))} \right| + \left| \frac{-L (g(x) - M)}{M(g(x))} \right| \ &\leq \frac{|M(f(x)) - L|}{M(g(x))} + \frac{|-L(g(x)-M)|}{M(g(x))} \ &= \frac{1}{M g(x)}|f(x)-L| + \frac{L}{M g(x)}|g(x) - M| \ &\leq \frac{2}{3} \end{align*}

\noindent \textbf{Comment:} The Reverse Triangle Inequality greatly simplifies the proof that [ \lim_{x\to a} g(x) = M \text{ implies } \lim_{x\to a} |g(x)| = |M|. ]

\noindent As $||g(x)| - |M|| \leq |g(x) - M|$, when $|g(x) - M| < \varepsilon$, we also have that $||g(x)| - |M|| < \varepsilon$.

\noindent \textbf{Definition/Notation:} Suppose we have $D \subseteq \mathbb{R}$ and $E \subseteq \mathbb{R}$ and functions [ F: D \to E \text{ and } g: D \to E. ]

\noindent We define a function $\left(\frac{f}{g}\right): D \setminus {x \in D ,|, g(x) = 0} \to E$ by:

[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \quad \text{ for all } x \in \text{Dom}\left(\frac{f}{g}\right) ]

\noindent \vspace{0.5em}

\noindent Our next ``goal" will be to show that: $\displaystyle \lim_{x\to a} \left(\frac{f}{g}\right)(x) = \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x\to a} f(x)}{\displaystyle\lim_{x\to a} g(x)} = \frac{L}{M}$

\noindent \textbf{Note:} In order for this to make sense, we will need $\displaystyle \lim_{x \to a} g(x) = M \neq 0$ (can’t divide by zero).

\begin{align*} & \text{``Scratch Work’’} \ \left| \frac{f(x)}{g(x)} - \frac{L}{M} \right| & = \left| \frac{M \cdot f(x) - L \cdot g(x)}{M \cdot g(x)} \right| \ & = \left| \frac{M \cdot f(x) - M \cdot L + M \cdot L - L \cdot g(x)}{M \cdot g(x)} \right| \ & = \left| \frac{M(f(x) - L) - L(g(x) - M)}{M \cdot g(x)} \right| \ & = \left| \frac{f(x)-L}{g(x)} + \frac{-L}{M g(x)} \cdot (g(x) - M) \right| \end{align*}

As we have $ \lim_{x \to a} f(x) = L $ and $ \lim_{x \to a} g(x) = M $, we try to get expressions of the form $|f(x) - L|$ and $|g(x) - M|$.

\begin{align*} \left| \frac{f(x)-L}{g(x)} + \frac{-L}{M g(x)} \cdot (g(x) - M) \right| & \leq \left| \frac{f(x)-L}{g(x)} \right| + \left| \frac{-L}{M g(x)} \cdot (g(x) - M) \right| \text{ Triangle Inequality} \ &= \frac{1}{|g(x)|} \left| f(x) - L \right| + \left| \frac{L}{M g(x)} \right| \left| g(x) - M \right|\ & \leq |A||f(x) - L| + |B||g(x) - M| \ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*} We already have $M \neq 0$, but we’ll need $|g(x)| \neq 0$ on some $\delta$-window around $a$. As $\lim_{x \to a} g(x) \neq 0$, this will be possible.