Limits of functions

Limits of functions at infinity

Definition

For a function \( f \), we say that \(\lim_{x \to +\infty} f(x) = L\) if for all \( \epsilon > 0 \) there is a number \( N > 0 \) such that \(|f(x) - L| < \epsilon \) for all \( x > N \).

Definition

For a function \( f \), we say that \(\lim_{x \to -\infty} f(x) = L\) if for all \( \epsilon > 0 \) there is a number \( N < 0 \) such that \(|f(x) - L| < \epsilon \) for all \( x < N \).

Definition

For a function \( f \), we say that \(\lim_{x \to +\infty} f(x) = \infty\) if for all \( M > 0 \) there is a number \( N > 0 \) such that \( f(x) > M \) for all \( x > N \).

Limits of functions at real numbers

Epsilon delta definition

For a function \( f \), we say that \(\lim_{x \to a} f(x) = L \) if for all \( \varepsilon > 0 \) there is a number \( \delta > 0 \) such that \( |f(x) - L| < \varepsilon \) for all \( x \) with \( 0 < |x - a| < \delta \).

Limits on the right and left of a point

Definitions

• We say that

  \[ \lim_{x \to a^-} f(x) = L \]

  if for all \(\varepsilon > 0\) there exists some \(\delta > 0\) such that if \(a - x < \delta\) for all \(x < a\) then \(|f(x) - L| < \varepsilon\).

• We say that

  \[ \lim_{x \to a^+} f(x) = L \]

  if for all \(\varepsilon > 0\) there exists some \(\delta > 0\) such that if \(x - a < \delta\) for all \(x > a\) then \(|f(x) - L| < \varepsilon\).

Theorem

A function $f$ admits a limit at a point $a \in \mathbb{R}$ if and only if $f$ admits a left limit and right limit at $a$ and they coincide. That is,

\[ \lim_{x \to a} f(x) \Leftrightarrow \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \]

Sum rule

Theorem

If $f$ and $g$ are functions such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then

\[ \lim_{x \to a} (f(x) + g(x)) = L + M \]

Proof

Let \(\epsilon>0\) and let \(\epsilon_{1}>0\) and \(\epsilon_{2}>0\) be values so that \(\epsilon=\epsilon_{1}+\epsilon_{2}\). Since \(\lim\limits_{x\to a}f(x)\) exists and is equal to \(L\) we know that there exists some \(\delta_{1}>0\) such that

\[\text{if }0 < p|x-a|<\delta_{2}\text{ then }|g(x)-M|<\epsilon_{2} \]

So if we define \(\delta=\min\{\delta_{1},\delta_{2}\}\), then if \(0<|x-a|<\delta\) we have that \(|x-a|\) is also smaller than both \(\delta_{1}\) and \(\delta_{2}\), so

\[|f(x)-L| < \epsilon_{1}\qquad\text{and}\qquad|g(x)-M| < \epsilon_{2} \]

We can now apply the triangle inequality to see that

$$ |(f+g)(x)-(L+M)| = |f(x)+g(x)-L-M| $$ $$= |(f(x)-L)+(g(x)-M)|$$ $$\leq |f(x)-L|+|g(x)-M|$$ $$< \epsilon_{1}+\epsilon_{2}$$ $$= \epsilon$$

So for any \(\epsilon>0\) we have found a value \(\delta>0\) such that

\[\text{if }0<|x-a|<\delta\text{ then }|(f+g)(x)-(L+M)|<\epsilon \]

$\square$

Product rule

Theorem

If $f$ and $g$ are functions such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then

\[ \lim_{x \to a} (f(x) \cdot g(x)) = L \cdot M \]

Proof

Let \(\epsilon>0\) be given. Since \(\lim_{x \to a} f(x) = L\), we can find a \(\delta_{1}>0\) such that if \(0<|x-a|<\delta_{1}\) then \(|f(x)-L|<\frac{\epsilon}{2(|M|+1)}\). Similarly, since \(\lim_{x \to a} g(x) = M\), we can find a \(\delta_{2}>0\) such that if \(0<|x-a|<\delta_{2}\) then \(|g(x)-M|<\frac{\epsilon}{2(|L|+1)}\).

So if we define \(\delta=\min\{\delta_{1},\delta_{2}\}\), then if \(0<|x-a|<\delta\) we have that \(|x-a|\) is also smaller than both \(\delta_{1}\) and \(\delta_{2}\), so

\[|f(x)-L| < \frac{\epsilon}{2(|M|+1)}\qquad\text{and}\qquad|g(x)-M| < \frac{\epsilon}{2(|L|+1)}\]

We can now apply the product limit theorem to see that

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

So for any \(\epsilon>0\) we have found a value \(\delta>0\) such that

\[\text{if }0<|x-a|<\delta\text{ then }|(f \cdot g)(x)-(L \cdot M)|<\epsilon \]

$\square$

Squeeze theorem, sandwich theorem

Theorem

Let \( f, g \), and \( h \) be functions such that

\[ f(x) \leq g(x) \leq h(x) \text{ for all } x \in \mathbb{R} \]

if

\[ \lim_{x \to a} f(x) = L \text{ and } \lim_{x \to a} h(x) = L \text{ then } \lim_{x \to a} g(x) = L \]

Proof

For all \(\epsilon > 0\) there exists \(\delta > 0\) such that

\[ \text{if } |x - a| < \delta \text{ then } |f(x) - L| < \epsilon, \text{ and} \] \[ \text{if } |x - a| < \delta \text{ then } |h(x) - L| < \epsilon \]

Recall that \(|f(x) - L| < \epsilon\) and \(|h(x) - L| < \epsilon\) is equivalent to writing

\[ -\epsilon < f(x) - L < \epsilon \quad \text{and} \quad -\epsilon < h(x) - L < \epsilon \]

So if \(|x - a| < \delta\) then

\[ -\epsilon < f(x) - L \leq g(x) - L \leq h(x) - L < \epsilon \]

hence

\[ -\epsilon \leq g(x) - L \leq \epsilon \]

So \(|g(x) - L| < \epsilon\)

$\square$