Section 6.8: Indeterminate Forms

Review

Trigonometric functions:

GraphicsGrid[{{Plot[Sin[x],{x,-10,10},PlotLabel->HoldForm[y=Sin[x]]]},{Plot[Tan[x],{x,-10,10},Exclusions->{-5π/2,-3π/2,-π/2,π/2,3π/2,5π/2},PlotLabel->HoldForm[y=Tan[x]]]},{Plot[Cos[x],{x,-10,10},PlotLabel->HoldForm[y=Cos[x]]]}
}]

Questions

What is the domain and range of (sin^(-1))(x)? Of (tan^(-1))(x)? Of (cos^(-1))(x)?

What is tan((tan^(-1))(0.3))? What is (cos^(-1))(cos(π/5))? What is sin((cos^(-1))(1/2))?

Tan[ArcTan[0.3]]
ArcCos[Cos[Pi/5]]
Sin[ArcCos[1/2]]

Right triangles

To find tan((sin^(-1))(0.4)), let θ=(sin^(-1))(0.4) so that 0.4=sin(θ). Represent sin(θ) in a right triangle.

Graphics[{Line[{{0,0},{0.8,0},{0.8,0.4},{0,0}}],Circle[{0,0},0.12,{0,ArcSin[0.4]}],Text[θ,0.15{Cos[.5ArcSin[0.4]],Sin[.5ArcSin[0.4]]}],Text[0.4,{0.85,0.2}],Text[1,{0.4,0.25}]}]

Using this triangle, tan((sin^(-1))(0.4))=tan(θ)=(0.4/(Sqrt[1-(0.4^2)]))=(0.4/(Sqrt[0.84]))≈0.436.

Questions

Use a right triangle to find a formula for sec((tan^(-1))(x)).

Simplify[Sec[ArcTan[x]]]

Indeterminate forms

Questions

What is (lim_(x->0))((2x+1)/(x-1))?

Limit[(2x+1)/(x-1),x->0]

What is (lim_(x->0))((4x+1)/x)?

Limit[(4x+1)/x,x->0]

What is (lim_(x->0))((2x)/x)?

Limit[2x/x,x->0]

What is (lim_(x->0))(x/(5x))?

Limit[x/(5x),x->0]

What is (lim_(x->∞))(x/2)?

Limit[x/2,x->Infinity]

What is (lim_(x->∞))(1/x)?

Limit[1/x,x->Infinity]

What is (lim_(x->-∞))((x-1)/(x+1))?

Limit[(x-1)/(x+1),x->-Infinity]

What is (lim_(x->∞))((x^2)/x)?

Limit[x^2/x,x->Infinity]

Given a limit (lim_(x->a))f(x), if we can simply evaluate f(a) as the limit we say (lim_(x->a))f(x) is determinate (and continuous). If we cannot simply evaluate f(x) at x=a, we say the limit is indeterminate -- it may or may not exist. Perhaps some algebra will help....

Indeterminate limits

(0/0), a small number divided by a small number could be anything. More work is needed.

(∞/∞), a large number divided by a large number could be anything. More work is needed.

L’Hopital’s rule

If you have a limit of a quotient which is either a (0/0) or an (∞/∞) limit, then the following is true if the limit (and the derivatives) exists:
	(lim_(x->a))((g(x))/(h(x)))=(lim_(x->a))((g'(x))/(h'(x)))
Warning: If the limit is not (0/0) or (∞/∞), the above two limits are not equal.

Example

To evaluate (lim_(x->∞))(x/(e^x)), first notice that plugging in ∞ for x produces (∞/(e^∞))=(∞/∞). We can use L’Hopital’s rule:
	(lim_(x->∞))(x/(e^x))=(lim_(x->∞))((((x))')/((((e^x)))'))=(lim_(x->∞))(1/(e^x))=0

Questions

Evaluate the following limits.

(lim_(x->∞))(x/(ln(x)))

Limit[x/Log[x],x->Infinity]

(lim_(x->∞))(((tan^(-1))(x))/((x^2)+1))

Limit[(ArcTan[x])/(x^2+1),x->Infinity]

(lim_(x->1))(((x^2)-3x+2)/(sin(x-1)))

Limit[(x^2-3x+2)/(Sin[x-1]),x->1]

(lim_(x->0))((1-cos(x))/(x^2))

Limit[(1-Cos[x])/x^2,x->0]

Why it works

For the (0/0) case, this means f(a)=0 and g(a)=0. Remember the limit definition of the derivative
	f'(a)=(lim_(x->a))((f(x)-f(a))/(x-a))
	g'(a)=(lim_(x->a))((g(x)-g(a))/(x-a))
Since f(a)=0=g(a)
	(lim_(x->a))((f(x))/(g(x)))=(lim_(x->a))((f(x)-f(a))/(g(x)-g(a)))
	=(lim_(x->a))(((f(x)-f(a))/(x-a))/((g(x)-g(a))/(x-a)))
	=((f'(a))/(g'(a)))

Questions

Can we use L’Hopital’s rule on (lim_(x->1))((x-1)/(e^(x-1)))? Compare the actual value of this limit with the limit that comes from L’Hopital’s rule.

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

Other indeterminate forms

There are other indeterminate forms. Sometimes we can use L’Hopital’s rule to evaluate them if we can rewrite into either the (0/0) or (∞/∞) form.

Other forms

∞-∞ or a large number minus a large number.

0·∞ or a number close to zero times a large number.

Indeterminate powers

(0^0) or a small number raised to another small number.

(∞^0) or a large number raised to a small number.

(1^∞) or a number close to 1 raised to a large power.

Difference example

Evaluate (lim_(x->∞))(ln(2x+1)-ln(3x-5)).

Use log properties to rewrite as a fraction.

(lim_(x->∞))(ln(2x+1)-ln(3x-5))
	=(lim_(x->∞))ln(((2x+1)/(3x-5)))
	=ln((lim_(x->∞))((2x+1)/(3x-5)))  -- (What allows us to do this?)
	=?

Limit[Log[(2x+1)/(3x-5)],x->Infinity]

Product example

Evaluate (lim_(x->∞))x(π/2-(tan^(-1))(x)).

Rewrite one of the factors as a fraction, factor=(1/(1/factor)).

(lim_(x->∞))x(π/2-(tan^(-1))(x))=(lim_(x->∞))(((π/2-(tan^(-1))(x)))/(1/x))=?

Limit[x(Pi/2-ArcTan[x]),x->Infinity]

Questions

(lim_(x->∞))x sin((2/x))

Limit[x Sin[2/x],x->Infinity]

(lim_(x->(SuperPlus[0])))x ln(x)

Limit[x Log[x],x->0,Direction->"FromAbove"]

Power example

A limit that comes from finance is
	(lim_(x->∞))(((1+(r/x)))^x)

Change the exponential expression to base e and take the limit of the exponent.

(lim_(x->∞))(((1+(r/x)))^x)=(lim_(x->∞))(e^(ln((((1+(r/x)))^x))))=(lim_(x->∞))(e^(x ln(1+(r/x))))
	=(e^((lim_(x->∞))x ln(1+r/x)))=?

E^Limit[x Log[1+r/x],x->Infinity]

Questions

Consider the function g(x) = (((x + (e^x)))^(1/x)). It is defined on (0,∞).

What is (lim_(x->(SuperPlus[0])))g(x)?

Limit[(x+E^x)^(1/x),x->0,Direction->"FromAbove"]

What is (lim_(x->∞))g(x)?

Limit[(x+E^x)^(1/x),x->Infinity]