Calculus: The Language of Change

Exercise Set 5.1
The Gap for Power Functions

by Sharon Portz

Exercise 1 : Computing the Limit

y = (x^p) ⇒ dy = p (x^(p - 1))dx,  p = 1, 2, 3, ...
For each f[x] = (x^p) below:

a) Compute \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) and simplify.

b) Compute f'[x] = (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x Give an intuitive justification why your limit is correct.  Does x need to be bounded, or can it vary arbitrarily as Δx tends to zero?

c) Use your limit f'[x] to solve for ϵ and write the increment equation:

f[x + Δx] - f[x] = f'[x] Δx + ϵ Δx
		 = [term in x but not Δx]Δx + [observed microscopic error]Δx

Notice that we can solve the increment equation for ϵ = \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)- f'[x]

d) Show that ϵ → 0 as Δx → 0.  "Show" this in any way that you consider reasonable.  Does x need to be bounded, or can it vary arbitrarily as Δx tends to zero?

1) If f[x] = (x^1), then f'[x] = 1 (x^0) = 1 and ϵ = 0.

2) If f[x] = (x^2), then f'[x] = 2 (x^1) and ϵ = Δx.

3) If f[x] = (x^3), then f'[x] = 3 (x^2) and ϵ = (3x + Δx)Δx.

4) If f[x] = (x^4), then f'[x] = 4 (x^3) and ϵ = (6 (x^2) + 4x Δx + Δ(x^2))Δx.

5) If f[x] = (x^5), then f'[x] = 5 (x^4) and ϵ = (10 (x^3) + 10(x^2) Δx + 5xΔ(x^2) + Δ(x^3))Δx.

part a

a) Compute \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) and simplify.

Solution of part a (f[x] = (x^1))

\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)  	= \(TraditionalForm
      \`\(\((x\  + \ Δx)\)\^1\  - \ \ x\^1\ \)\/Δx
        \)

= \(TraditionalForm
      \`\(x\  + \ Δx\  - \ x\)\/Δx\)

= Δ(x/Δ)x

= 1

Solution of part a (f[x] = (x^2))

\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) = \(TraditionalForm
      \`\(\((x\  + \ Δx)\)\^2\  - \ \ x\^2\ \)\/Δx
        \)

= (((x^2)+2 x Δx+Δ(x^2) -  (x^2))/(Δx))

= \(TraditionalForm
      \`\(2  x\ Δx\  + \ Δx\^2\)\/Δx
        \)

= 2x + Δx

Solution of part a (f[x] = (x^3))

\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) = \(TraditionalForm
      \`\(\((x\  + \ Δx)\)\^3\  - \ \ x\^3\ \)\/Δx
        \)

= (((x^3)+3 (x^2) Δx+3 x Δ(x^2)+Δ(x^3)- (x^3))/(Δx))

= \(TraditionalForm
      \`\(3\ x\^2\ Δx + 3\ x\ Δx\^2 + 
          Δx\^3\)\/Δx\)

= 3(x^2) + 3xΔx + Δ(x^2)

Solution of part a (f[x] = (x^4))

\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) = \(TraditionalForm
      \`\(\((x\  + \ Δx)\)\^4\  - \ \ x\^4\ \)\/Δx
        \)

= \(TraditionalForm
      \`\(x\^4 + 4\ x\^3\ Δx + 6\ x\^2\ Δx\^2 + 
          4\ x\ Δx\^3 + Δx\^4\  - \ 
          x\^4\)\/Δx\)

= \(TraditionalForm
      \`\(4\ x\^3\ Δx + 6\ x\^2\ Δx\^2 + 
          4\ x\ Δx\^3 + Δx\^4\)\/Δx
        \)

= 4 (x^(3 )) +  6 (x^2) Δx+4 x Δ(x^2) + Δ(x^3)

Solution of part a (f[x] = (x^5))

\(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) = \(TraditionalForm
      \`\(\((x\  + \ Δx)\)\^5\  - \ \ x\^5\ \)\/Δx
        \)

= (((x^5)+5 (x^4) Δx+10 (x^3) Δ(x^2)+10 (x^2) Δ(x^3)+5 x Δ(x^4)+Δ(x^5) - (x^5))/(Δx))

= \(TraditionalForm
      \`\(5\ \(x\^4\) Δx + 10\ \(x\^3\) Δx\^2 + 
          10\ \(x\^2\) Δx\^3 + 5\ x\ Δx\^4 + 
          Δx\^5\)\/Δx\)

= 5 (x^4) + 10 (x^3) Δx +10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4)

part b

b) Compute f'[x] = (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x Give an intuitive justification why your limit is correct.  Does x need to be bounded, or can it vary arbitrarily  as Δx tends to zero?

Solution of part b (f[x] = (x^1))

f'[x] 	= (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x

= (lim_(Δx →   0)) 
      (((((x  +  Δx))^1)  -   (x^1) )/Δ)x

= (lim_(Δx →  0)) 1

= 1

An  intuitive  explanation :   
      (((((x  +  Δx))^1)  -   (x^1) )/Δ)x
         = 1 from part (a).  Since there  is no Δx in 1, the limit  as Δx goes to 0 of 1 is just 1.

There  is no x-term  here  so x can vary arbitrarily, and f'[x] is always 1.

Solution of part b (f[x] = (x^2))

f'[x] 	= (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x

= (lim_(Δx  → 0)) 
      (((((x  +  Δx))^2)  -   (x^2) )/Δ)x

= (lim_(Δx  →  0))2x + Δx

= 2x

Intuitive  explanation:  As Δx goes to 0 it drops out of the  expression 2x + Δx.

Here we have an x term and a separate Δx term. So f'[x] goes to 2x as Δx goes to 0 for any value of x.  But if x is not bounded, as x goes to infinity, f'[x] also goes to infinity.  So to bound f'[x] we need to bound x.

Solution of part b (f[x] = (x^3))

f'[x] 	= (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x

= (lim_(Δx  →  0)) 
      (((((x  +  Δx))^3)  -   (x^3) )/Δ)x

= (lim_(Δx  →  0)) 3(x^2) + 3xΔx + Δ(x^2)

= 3(x^2)

Intuitively this is true because as Δx gets closer to 0, 3xΔx gets closer to 0 and so doesΔ(x^2).  That  leaves 3(x^2).

In this case x must be bounded.  If x varied arbirarily  (x → ∞) then we would have ∞ * 0 which is not clearly defined so x must be bounded.

Solution of part b (f[x] = (x^4))

f'[x] 	= (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x

= (lim_(Δx  →  0)) 
      (((((x  +  Δx))^4)  -   (x^4) )/Δ)x

= \(TraditionalForm
      \`\(\(lim\_\(Δx\  → \ 0\)\ 4\ x\^\(3\ \) + 
        \)\ \)\)6 (x^2) Δx+4 x Δ(x^2) + Δ(x^3)

= 4(x^(3 ))

Intuitively:  Again, as Δx gets close to 0 it causes the entire  6 (x^2) Δx-term  to go to 0, the 4 x Δ(x^2)-term  to go to zero, and Δ(x^3) to go to zero.  This leaves 4 (x^3).

As in f[x] = (x^3), x must be bounded in this case as well.

Solution of part b (f[x] = (x^5))

f'[x] 	= (lim_(Δx  →  0))((f[x  +  Δx]  -  f[x])/Δ)x

= (lim_(Δx  →  0)) 
      (((((x  +  Δx))^5)  -   (x^5) )/Δ)x

= (lim_(Δx  →  0)) 5 (x^4) + 10 (x^3) Δx +10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4)

= 5 (x^4) + 10 (x^3) Δx +10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4)

= 5  (x^4)

Intuitively this is true because Δx → 0 causes the terms 10 (x^3) Δx,  10 (x^2) Δ(x^2),  5 x Δ(x^3), and Δ(x^4) to go to zero.

As in the previous two functions, x must be bounded.

part c

c) Use your limit f'[x] to solve for ϵ and write the increment  equation:

f[x + Δx] - f[x] = f'[x] Δx + ϵ Δx

= [term in x but not Δx]Δx + [observed microscopic error]Δx

Notice that we can solve the increment  equation for ϵ = \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)- f'[x]

Solution of part c (f[x] = (x^1))

From part (a) we have: \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\) = 1

From part (b) we have: f'[x] = 1

Putting these into:

ϵ  = \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)- f'[x]

we have:

ϵ = 1- 1

= 0

The increment  equation:

f[x + Δx] - f[x] = Δx

f'[x] Δx + ϵ Δx = 1Δx + ϵ 0

= Δx

Thus  f[x + Δx] - f[x] 	= f'[x] Δx + ϵ Δx

= Δx

Solution of part c (f[x] = (x^2))

From part (a) we have:	\(TraditionalForm
      \`\(\(\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx = 
        \)\ \)\)2x + Δx

From part (b) we have:	f'[x] = 2x

Thus,	ϵ = 2x + Δx - 2x

= Δx

The increment  equation:

f[x + Δx] - f[x]= 2x Δx + Δ(x^2)

f'[x] Δx + ϵ Δx = 2x Δx + Δx Δx

= 2x Δx + Δ(x^2)

Thus  f[x + Δx] - f[x] 	= f'[x] Δx + ϵ Δx

= 2x Δx + Δ(x^2)

Solution of part c (f[x] = (x^3))

From part (a) we have:	\(TraditionalForm
      \`\(\(\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx = 
        \)\ \)\)3(x^2) + 3xΔx + Δ(x^2)

From part (b) we have:	f'[x] = 3(x^2)

Thus,	ϵ = 3(x^2) + 3xΔx + Δ(x^2) - 3(x^2)

= 3xΔx + Δ(x^2)

The increment  equation:

f[x + Δx] - f[x] = 3(x^2)Δx + 3xΔ(x^2) + Δ(x^3)

f'[x] Δx + ϵ Δx = 3(x^2)Δx + (3xΔx + Δ(x^2))Δx

= 3(x^2)Δx + 3xΔ(x^2) + Δ(x^3)

Thus  f[x + Δx] - f[x] 	= f'[x] Δx + ϵ Δx

= 3(x^2)Δx + 3xΔ(x^2) + Δ(x^3)

Solution of part c (f[x] = (x^4))

From part (a) we have:	\(TraditionalForm
      \`\(\(\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx = 
        \)\ \)\)4 (x^(3 )) +  6 (x^2) Δx+4 x Δ(x^2) + Δ(x^3)

From part (b) we have:	f'[x] = 4(x^(3 ))

Thus,	ϵ = 4 (x^(3 ))+ 6 (x^2) Δx+4 x Δ(x^2)+ Δ(x^3) - 4 (x^(3 ))

= 6 (x^2) Δx+4 x Δ(x^2)+ Δ(x^3)

The increment  equation:

f[x + Δx] - f[x] = 4 (x^(3 )) Δx +  6 (x^2) Δ(x^2) +4 x Δ(x^3) + Δ(x^4)

f'[x] Δx + ϵ Δx = 4 (x^(3 )) Δx + (6 (x^2) Δx+4 x Δ(x^2)+ Δ(x^3))Δx

= 4 (x^(3 )) Δx + 6 (x^2) Δ(x^2) + 4 x Δ(x^3) + Δ(x^4)

Thus  f[x + Δx] - f[x] 	= f'[x] Δx + ϵ Δx

= 4 (x^(3 )) Δx +  6 (x^2) Δ(x^2) +4 x Δ(x^3) + Δ(x^4)

Solution of part c (f[x] = (x^5))

From part (a) we have:	\(TraditionalForm
      \`\(\(\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx = 
        \)\ \)\) 5 (x^4) + 10 (x^3) Δx +10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4)

From part (b) we have:	f'[x] = 5  (x^4)

Thus,	ϵ =  5 (x^4) + 10 (x^3) Δx+10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4) - 5  (x^4)

= 10 (x^3) Δx+10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4)

The increment  equation:

f[x + Δx] - f[x] = 5 (x^4)Δx + 10 (x^3) Δ(x^2) +10 (x^2) Δ(x^3) + 5 x Δ(x^4) + Δ(x^5)

f'[x] Δx + ϵ Δx = 5 (x^4)Δx + (10 (x^3) Δx+10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4))Δx

=  5 (x^4)Δx + 10 (x^3) Δ(x^2) +10 (x^2) Δ(x^3) + 5 x Δ(x^4) + Δ(x^5)

Thus  f[x + Δx] - f[x] 	= f'[x] Δx + ϵ Δx

= 5 (x^4)Δx + 10 (x^3) Δ(x^2) +10 (x^2) Δ(x^3) + 5 x Δ(x^4) + Δ(x^5)

part d

d) Show that ϵ → 0 as Δx → 0.  "Show" this in any way that you consider reasonable.  Does x need to be bounded, or can it vary arbitrarily as Δx tends to zero?

Solution of part d (f[x] = (x^1))

From part (c) we have

ϵ = 0

There  is nothing to show.

Solution of part d (f[x] = (x^2))

From part (c) we have

ϵ = Δx

So as Δx → 0 then ϵ must also go to 0.

Solution of part d (f[x] = (x^3))

From part (c) we have
	ϵ = 3xΔx + Δ(x^2)

= (3 x + Δx)Δx

So as Δx gets closer to 0, the product that  remains , (3x + 0)0 , also goes to zero.  You can see this more clearly by setting x equal to a value and plugging in values for Δx.  See below.

Set x = 15 
Let Δx = 1
	ϵ = (3*15 + 1) 1
	   = 46

Let Δx = 0.5
	ϵ = (3*15 + 0.5)0.5
	   = 22.75

Let Δx = 0.00005
	ϵ = ( 3*15 + 0.00005) 0.00005
	   = 0.00225

x must be bounded.  If it were not, we would get something like this:

Let x = 100, Δx = 0.05
	ϵ = (3*100 + 0.05)0.05
	   = 15.0025

Let x = 58942, Δx = 0.005
	ϵ =  (3*58942 + 0.005)0.005
	   = 884.13

Solution of part d (f[x] = (x^4))

From part (c) we have
	ϵ = 6 (x^2) Δx+4 x Δ(x^2)+ Δ(x^3)
	   = Δx (6 (x^2) + 4x Δx + Δ(x^2))

So as Δx gets closer to 0, the product that  remains ,0(6 (x^2) + 4x 0 + 0) , also goes to zero.  You can see this more clearly by setting x equal to a value and plugging in values for Δx.

Use the input cell below to run calculations on your own.  Just set x equal to a number and vary Δx.

Clear[x,  Δx]x = 
Δx = 
Evaluate[Δx (6 (x^2) + 4  x Δx + Δ(x^2))]

For similar reasons shown above for f[x] = (x^3), x must be bounded here as well.

Solution of part d (f[x] = (x^5))

From part (c) we have
	ϵ = 10 (x^3) Δx+10 (x^2) Δ(x^2) + 5 x Δ(x^3) + Δ(x^4)
	   = Δx (10 (x^3) + 10 (x^2) Δx + 5 x Δ(x^2) + Δ(x^3))

So as Δx gets closer to 0, the product that  remains , 0(10 (x^3) + 10 (x^2) 0 + 5 x 0 + 0) , also goes to zero.  You can see this more clearly by plugging in values for Δx. Use the cell below to check on your own.

Clear[x,  Δx]x = 
Δx = 
Evaluate[Δx (10 (x^3) + 10 (x^2) Δx + 5 x Δ(x^2) + Δ(x^3))]

For similar reasons shown above for f[x] = (x^3), x must be bounded here as well.