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Week 10 Discussion Notes

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Limits of Functions

Definition

Given a function f ⁣:D→R\func{f}{D}{\R}, we say that lim⁡x→af(x)=L\displaystyle\lim_{x\to a} f\p{x} = L if for every sequence {xn}n⊆D∖{a}\set{x_n}_n \subseteq D \setminus \set{a} such that xn→ax_n \to a, we have lim⁡n→∞f(xn)=L\displaystyle\lim_{n\to\infty} f\p{x_n} = L.

Compare with the definition of continuity:

f ⁣:D→R\func{f}{D}{\R} is continuous at a∈Da \in D if for every sequence {xn}n⊆D\set{x_n}_n \subseteq D such that xn→ax_n \to a, we have lim⁡n→∞f(xn)=L\displaystyle\lim_{n\to\infty} f\p{x_n} = L.

The difference here is that in continuity, the terms of the sequence are allowed to be aa, but in the definition of a limit, the elements of your sequence are not allowed to be aa. Beyond that, there's not much else to say. Everything is defined in terms of sequences, so as you would expect, all the usual limit laws apply.

Example 1.

Show that lim⁡x→1(x1/3−1x−1)(x2+1)=23\displaystyle\lim_{x\to1} \p{\frac{x^{1/3} - 1}{x - 1}}\p{x^2 + 1} = \frac{2}{3}.
Solution.

Recall that a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = \p{a - b}\p{a^2 + ab + b^2}. If we set a=x1/3a = x^{1/3} and b=1b = 1, then

x−1=(x1/3)3−1=(x1/3−1)(x2/3+x1/3+1),x - 1 = \p{x^{1/3}}^3 - 1 = \p{x^{1/3} - 1}\p{x^{2/3} + x^{1/3} + 1},

so

x1/3−1x−1=x1/3−1(x1/3−1)(x2/3+x1/3+1)=1x2/3+x1/3+1  ⟹  lim⁡x→1x1/3−1x−1=lim⁡x→11x2/3+x1/3+1=13.\begin{gathered} \frac{x^{1/3} - 1}{x - 1} = \frac{x^{1/3} - 1}{\p{x^{1/3} - 1}\p{x^{2/3} + x^{1/3} + 1}} = \frac{1}{x^{2/3} + x^{1/3} + 1} \\ \implies \lim_{x\to1} \frac{x^{1/3} - 1}{x - 1} = \lim_{x\to1} \frac{1}{x^{2/3} + x^{1/3} + 1} = \frac{1}{3}. \end{gathered}

Since polynomials are continuous, we also have lim⁡x→1(x2+1)=2\displaystyle\lim_{x\to1} \p{x^2 + 1} = 2, so by the limit laws,

lim⁡x→1(x1/3−1x−1)(x2+1)=(lim⁡x→1x1/3−1x−1)(lim⁡x→1(x2+1))=13⋅2=23.\lim_{x\to1} \p{\frac{x^{1/3} - 1}{x - 1}}\p{x^2 + 1} = \p{\lim_{x\to1} \frac{x^{1/3} - 1}{x - 1}} \p{\lim_{x\to1} \p{x^2 + 1}} = \frac{1}{3} \cdot 2 = \frac{2}{3}.

Derivatives

Definition

Let I⊆RI \subseteq \R be an interval and f ⁣:I→R\func{f}{I}{\R} be a function. Then ff is differentiable at a∈Ia \in I if

f′(a)=lim⁡x→af(x)−f(a)x−af'\p{a} = \lim_{x\to a} \frac{f\p{x} - f\p{a}}{x - a}

exists and is finite. In this case, we say f′(a)f'\p{a} is the derivative of ff at aa.

Here are some very useful formulas for calculating derivatives:

Proposition

Suppose f,gf, g are differentiable at a∈Ra \in \R and let c∈Rc \in \R. Then

  1. cfcf is differentiable at aa and (cf)′(a)=cf′(a)\p{cf}'\p{a} = cf'\p{a}.
  2. f+gf + g is differentiable at aa and (f+g)′(a)=f′(a)+g′(a)\p{f + g}'\p{a} = f'\p{a} + g'\p{a}.
  3. (product rule) fgfg is differentiable at aa and (fg)′(a)=f′(a)g(a)+f(a)g′(a)\p{fg}'\p{a} = f'\p{a}g\p{a} + f\p{a}g'\p{a}.
  4. (quotient rule) If g(a)≠0g\p{a} \neq 0, then fg\frac{f}{g} is differentiable at aa and (fg)′(a)=f′(a)g(a)−f(a)g′(a)[g(a)]2\p{\frac{f}{g}}'\p{a} = \frac{f'\p{a}g\p{a} - f\p{a}g'\p{a}}{\br{g\p{a}}^2}.
Theorem (chain rule)

Let a∈Ra \in \R. If ff is differentiable at g(a)g\p{a} and gg is differentiable at aa, then f∘gf \circ g is differentiable at aa and (f∘g)′(a)=f′(g(a))g′(a)\p{f \circ g}'\p{a} = f'\p{g\p{a}} g'\p{a}.

The rest of the notes will be dedicated to examples, since there's not much else to talk about.

Example 2.
(28.8)

Let f(x)=x2f\p{x} = x^2 for xx rational and f(x)=0f\p{x} = 0 for xx irrational. Show that ff is differentiable at x=0x = 0.
Solution.

First, let's look at a wrong answer:

Notice that f′(x)=2xf'\p{x} = 2x if xx is rational. Thus, f′(0)=0f'\p{0} = 0.

The problem with this statement is that ff is not continuous if x≠0x \neq 0, which means that f′(x)f'\p{x} doesn't exist for x≠0x \neq 0. As a result, "f′(x)f'\p{x} if xx is rational" is meaningless. This tells us that our formulas above won't be useful for this problem, and we'll have to do it directly from the definition.

We want to calculate lim⁡x→0f(x)−f(0)x−0\lim_{x\to0} \frac{f\p{x} - f\p{0}}{x - 0}, so it'll be a good idea to see what the function in the limit looks like. If x≠0x \neq 0, then

f(x)−f(0)x−0=f(x)x={xif x is rational,0otherwise.\frac{f\p{x} - f\p{0}}{x - 0} = \frac{f\p{x}}{x} = \begin{cases} x & \text{if } x \text{ is rational}, \\ 0 & \text{otherwise}. \end{cases}

This limit should be 00, so let's try to prove that. Let {xn}n\set{x_n}_n be a sequence such that xn≠0x_n \neq 0 for all nn and xn→0x_n \to 0. Let ε>0\epsilon > 0, so by definition, there exists N∈RN \in \R such that if n>Nn > N, then ∣xn∣<ε\abs{x_n} < \epsilon. Thus,

∣f(xn)−f(0)xn−0∣={∣xn∣if xn is rational,0otherwise<ε,\begin{aligned} \abs{\frac{f\p{x_n} - f\p{0}}{x_n - 0}} &= \begin{cases} \abs{x_n} & \text{if } x_n \text{ is rational}, \\ 0 & \text{otherwise} \end{cases} \\ &< \epsilon, \end{aligned}

since it's less than ε\epsilon in both cases. Thus, because {xn}n\set{x_n}_n was an arbitrary sequence,

f′(0)=lim⁡x→0f(x)−f(0)x−0=0.f'\p{0} = \lim_{x\to0} \frac{f\p{x} - f\p{0}}{x - 0} = 0.
Example 3.
(28.14)

Suppose ff is differentiable at aa. Prove:

  1. lim⁡h→0f(a+h)−f(a)h=f′(a)\displaystyle \lim_{h\to0} \frac{f\p{a + h} - f\p{a}}{h} = f'\p{a}
  2. lim⁡h→0f(a+h)−f(a−h)2h=f′(a)\displaystyle \lim_{h\to0} \frac{f\p{a + h} - f\p{a - h}}{2h} = f'\p{a}
Solution.

  1. Since ff is differentiable at aa, we know that for any sequence {xn}n\set{x_n}_n such that xn≠ax_n \neq a and xn→ax_n \to a, we have

    f′(a)=lim⁡n→∞f(xn)−f(a)xn−a.f'\p{a} = \lim_{n\to\infty} \frac{f\p{x_n} - f\p{a}}{x_n - a}.

    Let {hn}n\set{h_n}_n be a sequence such that hn≠0h_n \neq 0 and hn→0h_n \to 0. Notice that a+hn≠aa + h_n \neq a for all nn and a+hn→0a + h_n \to 0, so if we set xn=a+hnx_n = a + h_n, we can apply the equation above to get

    lim⁡n→∞f(a+hn)−f(a)hn=lim⁡n→∞f(a+hn)−f(a)(a+hn)−a=lim⁡n→∞f(xn)−f(a)xn−a=f′(a).\begin{aligned} \lim_{n\to\infty} \frac{f\p{a + h_n} - f\p{a}}{h_n} &= \lim_{n\to\infty} \frac{f\p{a + h_n} - f\p{a}}{\p{a + h_n} - a} \\ &= \lim_{n\to\infty} \frac{f\p{x_n} - f\p{a}}{x_n - a} \\ &= f'\p{a}. \end{aligned}

    Since {hn}n\set{h_n}_n was an arbitrary sequence, it follows that

    lim⁡h→0f(a+h)−f(a)h=f′(a).\lim_{h\to0} \frac{f\p{a + h} - f\p{a}}{h} = f'\p{a}.

  2. Notice that

    f(a+h)−f(a−h)2h=f(a+h)−f(a)+f(a)−f(a−h)2h=12(f(a+h)−f(a)h+f(a)−f(a−h)h)=12(f(a+h)−f(a)h+f(a+(−h))−f(a)(−h)).\begin{aligned} \frac{f\p{a + h} - f\p{a - h}}{2h} &= \frac{f\p{a + h} - f\p{a} + f\p{a} - f\p{a - h}}{2h} \\ &= \frac{1}{2} \p{\frac{f\p{a + h} - f\p{a}}{h} + \frac{f\p{a} - f\p{a - h}}{h}} \\ &= \frac{1}{2} \p{\frac{f\p{a + h} - f\p{a}}{h} + \frac{f\p{a + \p{-h}} - f\p{a}}{\p{-h}}}. \end{aligned}

    We can calculate the limit of both terms in the parentheses by the definition of the derivative, so by the limit laws,

    lim⁡h→0f(a+h)−f(a−h)2h=lim⁡h→012(f(a+h)−f(a)h+f(a+(−h))−f(a)(−h))=12(f′(a)+f′(a))=f′(a).\begin{aligned} \lim_{h\to0} \frac{f\p{a + h} - f\p{a - h}}{2h} &= \lim_{h\to0} \frac{1}{2} \p{\frac{f\p{a + h} - f\p{a}}{h} + \frac{f\p{a + \p{-h}} - f\p{a}}{\p{-h}}} \\ &= \frac{1}{2} \p{f'\p{a} + f'\p{a}} \\ &= f'\p{a}. \end{aligned}
Example 4.

True or false: If lim⁡h→0f(a+h)−f(a−h)2h\displaystyle \lim_{h\to0} \frac{f\p{a + h} - f\p{a - h}}{2h} exists and is finite, then ff is differentiable at aa.
Solution.

False. The exercise above tells us that if the derivative exists, then we can calculate f′(a)f'\p{a} using the above formula. If the derivative doesn't exist, then the formula may no longer be true. For example, if f(x)=∣x∣f\p{x} = \abs{x}, then

lim⁡h→0f(a+h)−f(a−h)2h=lim⁡h→0∣h∣−∣−h∣2h=0,\lim_{h\to0} \frac{f\p{a + h} - f\p{a - h}}{2h} = \lim_{h\to0} \frac{\abs{h} - \abs{-h}}{2h} = 0,

but ff is not differentiable at x=0x = 0.
Example 5.
(28.16)

Let ff be a function defined on an open interval II containing aa. Show that f′(a)f'\p{a} exists if and only if there is a function ε(x)\epsilon\p{x} defined on II and a real number bb such that

f(x)−f(a)=(x−a)(b−ε(x))andlim⁡x→aε(x)=0.f\p{x} - f\p{a} = \p{x - a}\p{b - \epsilon\p{x}} \quad\text{and}\quad \lim_{x\to a} \epsilon\p{x} = 0.
Solution.

"  ⟹  \implies"

Suppose f′(a)f'\p{a} exists. Let's try to reverse engineer what ε\epsilon and bb would have to be:

f(x)−f(a)=(x−a)(b−ε(x))  ⟹  ε(x)=b−f(x)−f(a)x−a.f\p{x} - f\p{a} = \p{x - a}\p{b - \epsilon\p{x}} \implies \epsilon\p{x} = b - \frac{f\p{x} - f\p{a}}{x - a}.

For ε(x)\epsilon\p{x} to tend to 00 as x→0x \to 0, we would need b=f′(a)b = f'\p{a}, by the definition of the derivative. Now let's prove that this works:

Technically, ε\epsilon is defined on I∖{a}I \setminus \set{a}, but its value at x=ax = a doesn't matter since we're taking limits, so we can set it to be any number we like to get a function defined on II. Then

f(x)−f(a)=(x−a)(f′(a)−ε(x))f\p{x} - f\p{a} = \p{x - a}\p{f'\p{a} - \epsilon\p{x}}

just by definition of ε(x)\epsilon\p{x}, and by our choice of bb, we get lim⁡x→aε(x)=0\displaystyle\lim_{x\to a} \epsilon\p{x} = 0 automatically.

"  ⟸  \impliedby"

Suppose that ε(x)\epsilon\p{x} and bb exist as in the problem statement. By rearranging the equation in the problem statement and applying limit laws,

f(x)−f(a)x−a=b−ε(x)  ⟹  lim⁡x→af(x)−f(a)x−a=lim⁡x→a(b−ε(x))=b−lim⁡x→aε(x)=b,\begin{aligned} \frac{f\p{x} - f\p{a}}{x - a} &= b - \epsilon\p{x} \\ \implies \lim_{x\to a} \frac{f\p{x} - f\p{a}}{x - a} &= \lim_{x\to a} \p{b - \epsilon\p{x}} \\ &= b - \lim_{x\to a} \epsilon\p{x} \\ &= b, \end{aligned}

so by definition of the derivative, f′(a)f'\p{a} exists and is equal to bb.