Calculus: The Language of Change

Exercise Set 8.3
Growth of Log, Exp, and Powers

by Sharon Portz

Use the program Infinities to explore orders of infinity and compute limits by computer.

Exercise 1 : Ratios of Powers

Show that  each of the functions g[x] tends to infinity faster than the corresponding f[x].

part a

g[x] = (x^5)

f[x] = (x^3)

Solution of part a

For each part of this problem we find the limit of the ratio of g[x]/f[x]as x goes to infinity.  Since the limit is infinity in each case, g[x] goes to infinity faster than f[x].

Clear[f, g, x]; f[x_]:=(x^3);
g[x_]:=(x^5);
Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

part b

g[x] = (x^2)

f[x] = (Sqrt[x])

Solution of part b

Clear[f, g, x]; f[x_]:=(Sqrt[x]);
g[x_]:=(x^2);
Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

part c

g[x] = (Sqrt[x])

f[x] = (x^(1/3))

Solution of part c

Clear[f, g, x]; f[x_]:= (x^(1/3));
g[x_]:=(Sqrt[x]);
Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

part d

g[x] = ((x^5)/10)

f[x] = 10 (x^3)

Solution of part d

Clear[f, g, x]; f[x_]:=10 (x^3);
g[x_]:=((x^5)/10);
Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

part e

g[x] = ((x^2)/2)

f[x] = 2 (Sqrt[x])

Solution of part e

Clear[f, g, x]; f[x_]:= 2 (Sqrt[x]);
g[x_]:=((x^2)/2);
Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

part f

g[x] = ((Sqrt[x])/2)

f[x] = 3 (x^(1/3))

Solution of part f

Clear[f, g, x]; f[x_]:= 3 (x^(1/3));
g[x_]:= ((Sqrt[x])/2);
Limit[g[x]/f[x], x -> ∞]Limit[f[x]/g[x], x -> ∞]

Exercise 2 : Order of Powers

Let f[x] = (Subscript[a, 1])(Subscript[(x^p), 1])+ (Subscript[b, 1])(Subscript[(x^q), 1]) + (Subscript[c, 1])(Subscript[(x^r), 1]) and g[x] = (Subscript[a, 2])(Subscript[(x^p), 2])+ (Subscript[b, 2])(Subscript[(x^q), 2])+ (Subscript[c, 2])(Subscript[(x^r), 2]) for positive constants (Subscript[a, i]),(Subscript[b, i]),(Subscript[c, i]),(Subscript[p, i]), (Subscript[q, i]), (Subscript[r, i]).

a) Show that f[x] has a higher order of infinity than g[x] when the highest power of f[x] is greater than the highest power of g[x].

b) What is the limit of the ratio f[x]/g[x]?

c) What is the limit if the highest powers agree?

part a

a) Show that f[x] has a higher order of infinity than g[x] when the highest power of f[x] is greater than the highest power of g[x].

Solution of part a

For ratios of single term functions:

	(Underscript[Lim, x -> ∞]) (((a (x^p))/(b (x^q)))) 
       = (a/b) (Underscript[Lim, x -> ∞]) (x^(p - q))

so if  p > q, p-q > 0 and the limit is Infinity no matter what the positive values of  a and  b.

The limit of the ratio of sums of functions depends on the "leading" term.  When one term in the numerator grows fastest, the limit is still Infinity.  For example,

(Underscript[Lim, x -> ∞]) 
        (((2 (x^3) - 4 (x^2) + 5 x - 7)/(800 (x^2) + 900 x + 1000)))
           =(Underscript[Lim, x -> ∞]) ((2 x - 4 + (5/x) - (7/(x^2)))/(800 + (900/x) + (1000/(x^2))))
 = (1/400) (Underscript[Lim, x -> ∞]) 
        ((x - (1/200) + (5/(800 x)) - (7/(800 (x^2))))/(1 + 
            (900/(800 x)) + (1000/(800 (x^2))))) =(((BIG - (1/200) + tiny))/(400 (1 + tiny + teeny)))= BIG

The six problems in Exercise 1 were all examples of the validity of the statement in this problem (with f and g reversed in order).

So whenever the highest power of f is greater than the highest power of g, the limit of the ratio (f/g) (as x goes to infinity) will be infinity and therefore f[x] has a higher order of infinity than g[x]. This is true even if the coefficients of the lower power terms of g are much larger.

For example:

Clear[f, g, x]; f[x_]:=x^5 ;
g[x_]:=50000000000x^4 + 50000000000  x^3;
Limit[f[x]/g[x], x -> ∞]

part b

b) What is the limit of the ratiof[x]/g[x]?

Solution of part b

As stated in part a, the limit of the ratio is Infinity.

part c

c) What is the limit if the highest powers agree?

Solution of part c

(Underscript[Lim, x -> ∞]) (((a (x^p))/(b (x^q)))) 
       = (a/b) (Underscript[Lim, x -> ∞]) (x^(p - q))

so if  p = q, p-q = 0 and the limit is (a/b) since the limit of (x^0) is 1.

Exercise 3: Drill Limits

Compute the limits (p, k > 0)

part a

(lim_(x -> 0))(x^p)Log[x]

Solution of part a

From Example 8.7 we know that

	(lim_(x -> 0))xLog[x] = 0
	
Since (lim_(x -> 0)) (x^(p - 1)) is also 0, 

	(lim_(x -> 0))(x^p)Log[x] = 0

\(\(
Limit[x\ Log[x], x -> 0]\)\)

part b

(lim_(x -> ∞))(x^((1/x)))

Solution of part b

If we write x = (e^Log)[x] then we have

(lim_(x -> ∞)) (x^((1/x))) = (lim_(x -> ∞))  (e^(Log[x]/x)) = (e^0) = 1

\(\(
Limit[x\^\(1\/x\), x -> ∞]\)\)

part c

(lim_(Δx -> 0)) Δ(x^((1/Log)[Δx]))

Solution of part c

Again we may write Δx = (e^Log)[Δx]so

(lim_(Δx -> 0)) Δ(x^((1/Log)[Δx])) = (e^(Log[Δx]/Log[Δx])) = (e^1) = e

\(\(
Limit[Δx^\((1/Log[Δx])\), 
      Δx -> 0]\)\)

part d

(lim_(x -> ∞)) (x^p)Log[x]

Solution of part d

This limit is ∞ because the limit of (x^p) is ∞, and the limit of Log[x] is also ∞.

\(\(
Limit[x^p\ Log[x], x -> ∞]\)\)

part e

(lim_(δ -> ∞)) ((1  +  kδ)/δ)

Solution of part e

((1  +  kδ)/δ) can be written as (1/δ) + k

so	(lim_(δ -> ∞)) ((1  +  kδ)/δ) = (lim_(δ -> ∞))(1/δ) +  (lim_(δ -> ∞)) k

	= 0 + k = k

\(\(


    Limit[\(1\  + \ k\ δ\)\/δ, δ -> ∞]
      \)\)

part f

(lim_(δ -> 0)) \(TraditionalForm
      \`δ\^\(\(δx\  + \ k\)\/Log[δ]\)\)

Solution of part f

(lim_(δ -> 0)) \(TraditionalForm
      \`δ\^\(\(δx\  + \ k\)\/Log[δ]\)\) = (lim_(δ -> 0)) \(TraditionalForm
      \`e\^\(\(\((δx\  + \ k)\) Log[δ]\)\/Log[δ]\)\)

	= (lim_(δ -> 0)) (e^((δx  +  k))) = (e^k)

Limit[(δ^(((δ x  +  k)/Log)[δ])) ,δ -> 0]

Exercise 4:

Suppose b > a > 0.  Find (lim_(x -> ∞))((b^x)/(a^x))

Solution  to #4

Since b > a and both b and a are greater than 0, the numerator is going to infinity faster than the denominator.  Thus the limit is ∞.

\(\(
Limit[3\^x\/2\^x, x -> ∞]\)\)

Exercise 5: Negative? Exponentials

If the base of an exponential function f[x] = (b^x) satisfies 0 < b < 1, we think of this as a "negative" exponential  in the following sense:

a) Solve for a constant  k so that  (b^x) = (e^(kx )) for all x.

b) Show that  k from part (a)  is negative when 0 < b < 1.

c) Show that  (lim_(x -> ∞)) (b^x) = 0 when 0 < b < 1.

part a

a) Solve for a constant  k so that  (b^x) = (e^(kx )) for all x.

Solution of part a

Begin with the equation (b^x) = (e^(kx )). 

Then take the Log of both sides

Log[(b^x)] = Log[(e^(kx ))]

so		x Log[b] = k x Log[e]

x Log[b] = k x

k = Log[b]

part b

b) Show that  k from part (a)  is negative  when 0 < b < 1.

Solution of part b

Look at the graph of Log[b] below.  If 0 < b < 1 then Log[b] is negative.

part c

c) Show that  (lim_(x -> ∞)) (b^x) = 0 when 0 < b < 1.

Solution of part c

Let's get an intuitive  idea about this first.  Setting  0 < b < 1 makes b a fraction (for rational b).  What  happens to fractions  when you take  powers of them??  They get smaller.  For example,

((((1/2)))^2) = (1/4), ((((1/2)))^3) = (1/8),  ((((1/2)))^4) = (1/16)...

So intuitively, (lim_(x -> ∞)) (b^x) = 0 when 0 < b < 1 is a true  statement.

Now, recall that  (b^x) = (e^(kx )) where k = Log[b] is negative.

So we can write the limit as

	(lim_(x -> ∞)) (b^x)  =(lim_(x -> ∞)) (e^(Log[b] x))

Since Log[b] is negative

	(lim_(x -> ∞)) (b^x) = (lim_(x -> ∞)) (e^(-c x))  where c ≻ 0
	
	=  (lim_(x -> ∞))(1/(e^(c x))) = 0

\(\(
Limit[0.999^x, x -> ∞]\)\)

Exercise 6: Orders of Infinity

Use logs to prove the other half of the Orders of Infinity theorem above.  If we wish to show that (lim_(x -> ∞))((x^p)/Log)[x] = ∞, then change  variables by letting u = Log[x], so x = (e^u).

We show instead that

(lim_(u -> ∞))((((e^u))^p)/u) = ∞

by taking (p^th) roots.  Why does this establish the result?

Solution

Before we begin, let's be sure we know what we are proving.  Theorem 8.5 states that  if p is any positive real number, then

(Underscript[(lim ), t -> ∞]) ((e^t)/(t^p)) = ∞ 		and		(Underscript[(lim ), t -> ∞])((t^p)/Log)[t] = ∞.

This is equivalent to

(Underscript[(lim ), t -> ∞]) ((t^p)/(e^t)) = 0		and		(Underscript[(lim ),  t -> ∞]) Log[t]/(t^p) = 0.

What  we are to prove is that   	(Underscript[(lim ), t -> ∞])((t^p)/Log)[t] = ∞ .

From  what is given above,

	(Underscript[(lim ), t -> ∞])((t^p)/Log)[t] = ∞  can be written as (lim_(u -> ∞))((((e^u))^p)/u) = ∞.

If we take the (p^th)root of ((((e^u))^p)/u) we get

\(TraditionalForm\`e\^u\/\@u\%p\) = ((e^u)/(u^(1/p)))

Thus our function is in a form we know.  It is in the form ((e^t)/(t^p)) where t = u and p = 1/p (which is still a positive number).  We know

(Underscript[(lim ), t -> ∞]) ((e^t)/(t^p)) = ∞ 	so 	 (Underscript[(lim ), u -> ∞])((e^u)/(u^(1/p))) = ∞

For example, if we let x = u and p = 5

\(\(
Limit[E^x\/x\^\(1\/5\), x -> ∞]\)\)

Evaluate the cell below to see the graph.

Clear[f, x, g, p, q]; \(f[x_] := E^x\/x\^\(1\/5\); 
Plot[f[x], {x, 1, 5}]\)

Exercise 7: All Exponentials Beat All Powers

Suppose b > 1 and p > 0.  What are the values of the limits
		(lim_(x -> ∞))((b^x)/(x^p))	and	(lim_(x -> ∞))((x^p)/(b^x)) ?

Solution

Evaluate the cells below choosing your own numbers for b and p and see what happens.

\(Clear[f, b, \ p, g, x]; 
f[x_] := b\^x; 
g[x_] := x\^p; 
b = \ 30; 

    p\  = \  .5; 
Limit[f[x]\/g[x], x -> ∞]\)\(Limit[g[x]\/f[x], x -> ∞]
\)

Evaluate the cell below to see the graphs.

\(Clear[f, b, \ p, g, x]; 
f[x_] := b\^x; 
g[x_] := x\^p; 
b = \ 30; 

    p\  = \  .5; 


    Plot[{f[x], \ g[x]}, \ {x, \ 0, \ 5}, 
	
      PlotStyle 
        -> {{Blue, AbsoluteThickness[1]}, {Red, AbsoluteThickness[3]}}]; 
    
Plot[f[x]/g[x], {x, 1, 10}, 
      PlotStyle -> {{DarkGreen, AbsoluteThickness[2]}}]; 

    Plot[g[x]/f[x], {x, 0, 10}, 
      PlotStyle -> {{Blue, AbsoluteThickness[2]}}]; \)

In the first graph, f[x] = (b^x) is in blue and g[x] =  (x^p) is in red.

It makes intuitive  sense that  if the numerator  is heading to infinity faster than the denominator, the limit of the quotient is infinity.  (In other words, the value of the numerator  is larger than the value of the demoninator.  Whenever  this happens, we get quotient  values greater  than 1.  The bigger the numerator  is than  the  denominator,  the bigger the quotient  value.)  See the second graph for a visual.

If the denominator  is heading to infinity faster than the numerator,  the limit of the quotient is 0.  (When we divide by large numbers the value is small.)  The third  graph demonstrates  this.