12. Limits of functions

What does it mean for a function to tend to a limit?

\text{Let }A\subseteq\mathbb{R}\text{ and }f:A\to\mathbb{R}.\ \text{We say that }f(x)\text{ tends to the limit }\ell\text{ as }x\to a\ (\text{write }\lim_{x\to a}f(x)=\ell)\text{ iff } \forall \varepsilon>0\ \exists\delta>0\ \text{s.t.}\ \forall x\in A,\ 0<|x-a|<\delta\Rightarrow |f(x)-\ell|<\varepsilon.

What does it mean for a function to tend to a complex number

\text{Let }A\subseteq\mathbb{C}\text{ and let }f:A\to\mathbb{C}\text{ be a function. We say that }f(z)\text{ tends to the limit }\ell\text{ as }z\text{ tends to }z_0\text{ (write }\lim_{z\to z_0}f(z)=\ell\text{) iff } \forall \varepsilon>0\ \exists\delta>0\ \text{s.t.}\ \forall z\in A,\ 0<|z-z_0|<\delta\Rightarrow |f(z)-\ell|<\varepsilon.

How do we prove that if the limit of a function exists then it is unique?

Assume \lim_{x\to a} f(x) exists but is not unique, so

\lim_{x\to a} f(x) = \ell \quad\text{and}\quad \lim_{x\to a} f(x) = m \quad\text{with } \ell \neq m.

1. Set \varepsilon = \frac{|\ell - m|}{3} > 0.

2. By The definition of tending to a limit, there are \delta_1, \delta_2 > 0 such that
   

   • if 0<|x-a|<\delta_1 then |f(x)-\ell| < \varepsilon,

   • if 0<|x-a|<\delta_2 then |f(x)-m| < \varepsilon.

1. Let \delta=\min\left\lbrace{\delta_1,\delta_2}\right\rbrace and pick x with 0<|x-a|<\delta. Then both inequalities hold, so
   |\ell - m| = |(\ell - f(x)) + (f(x) - m)| \le |,\ell - f(x),| + |,f(x) - m,| < \varepsilon + \varepsilon = \frac{2}{3}|\ell - m| < |\ell - m|,
   which is impossible.

Therefore our assumption \ell \neq m is wrong, so the limit (if it exists) must be unique.

What is “f is continuous at a” equivalent to and how do we prove this?

\lim_{x\to a}f\left(x)=f(a)\right.

Proof

Assume f is continuous at a. So for every \epsilon>0 there is \delta>0 such that for all x\in A, if |x-a|<\delta then |f(x)-f(a)|<\epsilon. The limit definition only adds the condition x\neq a (i.e. 0<|x-a|<\delta), so the same \delta works. Hence \lim_{x\to a}f(x)=f(a).”

proving the reverse.

Assume \lim_{x\to a}f(x)=f(a). Then for every \epsilon>0 there is \delta>0 such that for all x\in A, if 0<|x-a|<\delta then |f(x)-f(a)|<\epsilon. For x=a we have |f(a)-f(a)|=0<\epsilon automatically, so for all x\in A with |x-a|<\delta the inequality holds. This is exactly the \epsilon!-!\delta definition of continuity at a

Limit rule: multiplying a function by a constant

If f(x)\to \ell \text{ as } x\to a and c\in\mathbb{R}, then

cf(x)\to c\ell \text{ as } x\to a.

Reason:

|cf(x)-c\ell| = |c|\cdot |f(x)-\ell|, so choose \delta so that |f(x)-\ell|<\varepsilon/|c|.

Case when c = 0 as well.

Sum of two function limits

If

f(x)\to \ell \quad\text{and}\quad g(x)\to m \text{ as } x\to a,

then

f(x)+g(x)\to \ell+m.

Reason:

Use triangle inequality:

|(f+g)-(\ell+m)| \le |f-\ell| + |g-m|.

Limit rule: product of two limits

If

f(x)\to \ell \quad\text{and}\quad g(x)\to m,

then

f(x)g(x)\to \ell m.

Reason.

1. Make g(x) bounded:
   |g(x)-m|<1 \Rightarrow |g(x)|<1+|m|.

2. Choose \delta so
   |f-\ell|<\frac{\varepsilon}{2(1+|m|)}, \quad |g-m|<\frac{\varepsilon}{2(|\ell|+1)}.

3. Then
   |fg-\ell m| \le |f-\ell||g| + |\ell||g-m| < \varepsilon.

Limit rule: quotients of functions

If

g(x)\to m \neq 0,

then

\frac{1}{g(x)} \to \frac{1}{m}.

Reason:

Ensure |g(x)-m|<|m|/2 so |g(x)|>|m|/2.

Then

\left|\frac{1}{g(x)} - \frac{1}{m}\right| = \frac{|g(x)-m|}{|g(x)||m|} < \varepsilon.

If f(x) is always greater than or equal to zero, what must its limit be and the proof of this.

If

f(x)\ge 0 \text{ for all } x,

and

f(x)\to \ell,

then

\ell\ge 0.

Reason:

Assume \ell<0.

Choose \varepsilon = |\ell|/2.

Then |f(x)-\ell|<\varepsilon forces f(x)<0 — contradiction

If one function is always ≤ another, what happens to the limits?

If

f(x)\le g(x) \text{ for all } x,

and

f(x)\to \ell,\quad g(x)\to m,

then

\ell \le m.

Reason:

Consider

h(x)=g(x)-f(x)\ge0.

Its limit must be ≥0 by previous rules

What’s the sandwich rule for limits of a function?

Assume f,g,h:A\to\mathbb{R} and f(x)\le g(x)\le h(x) for all x\in A.

If \lim_{x\to a}f(x)=\lim_{x\to a}h(x)=\ell,

then \lim_{x\to a}g(x)=\ell

Proof:

Let \varepsilon>0.

Since \lim_{x\to a}f(x)=\ell, there is \delta_1>0 such that 0<|x-a|<\delta_1\Rightarrow |f(x)-\ell|<\varepsilon, i.e. \ell-\varepsilon<f(x)<\ell+\varepsilon . Since \lim_{x\to a}h(x)=\ell, there is \delta_2>0 such that 0<|x-a|<\delta_2\Rightarrow \ell-\varepsilon<h(x)<\ell+\varepsilon. Let \delta=\min\{\delta_1,\delta_2\}. Then for 0<|x-a|<\delta we have \ell-\varepsilon<f(x)\le g(x)\le h(x)<\ell+\varepsilon, so |g(x)-\ell|<\varepsilon. Hence \lim_{x\to a}g(x)=\ell."

What does it mean to converge from the right?

\forall\epsilon>0,\exists\delta>0 such that \forall x\in A ifx\in\left(a,a+\delta\right) then \left|f\left(x\right)-l\right|<\epsilon

What does it mean for a function to converge from the left

\forall\epsilon>0,\exists\delta>0 Such that \forall x\in A if x\in\left(a-\delta,a\right) then \left|f\left(x\right)-l\right|<\epsilon

What does it mean for the limit of a function to be infinity as x→a?

\forall m>0,\exists\delta>0 Such that \forall x\in A if 0<\left|x-a\right|<\delta then f\left(x\right)>m

What does it mean for the limit of a function to be a as x→\infty ?

\forall\epsilon>0,\exists k>0 Such that \forall x\in A, if x>k then \left|f\left(x\right)-l\right|<\epsilon