# Indeterminate Forms for Functions

## Related Calculator: Limit Calculator

In the note Limits Involving Infinity we saw that it is pretty easy to evaluate lim_(x->0)(x+1)/x because since lim_(x->0)(x+1)=1 and lim_(x->0)x=0 then division of 1 by very small number will give very large number, and so lim_(x->0)(x+1)/x=oo.

But there are some cases when it is not so easy to calculate limit. In fact we can't state limit in general without knowing functions.

There are 4 cases.

Case 1. Suppose that lim_(x->a)f(x)=0 and lim_(x->a)g(x)=0. We need to find lim_(x->a)f(x)/g(x).

It is not very clear what is the limit of the ratio and does it exist at all. Very small number divided by the very small number can give anything.

Let f(x)=1/x and g(x)=1/x^2. Clearly lim_(x->oo)1/x=0 and lim_(x->oo)1/x^2=0. Then lim_(x->oo)(f(x))/(g(x))=lim_(x->oo)(1/x)/(1/x^2)=lim_(x->oo)x=oo.

Let f(x)=1/x^2 and g(x)=1/x. Clearly lim_(x->oo)1/x^2=0 and lim_(x->oo)1/x=0. Then lim_(x->oo)(f(x))/(g(x))=lim_(x->oo)(1/x^2)/(1/x)=lim_(x->oo)1/x=0.

Let f(x)=3/x and g(x)=1/x. Clearly lim_(x->oo)3/x=0 and lim_(x->oo)1/x=0. Then lim_(x->oo)(f(x))/(g(x))=lim_(x->oo)(3/x)/(1/x)=lim_(x->oo)3=3.

Let f(x)=x and g(x)=x^2. Clearly lim_(x->0)x=0 and lim_(x->0)x^2=0. Then lim_(x->oo)(f(x))/(g(x))=lim_(x->oo)x/(x^2)=lim_(x->0)1/x doesn't exist because lim_(x->0^+)1/x=oo and lim_(x->0^-)1/x=-oo so one-sided limits are not equal.

Therefore, in general case we can't find limit of ratio without knowing functions.

To characterize this special case we say that when lim_(x->a)f(x)=0 and lim_(x->a)g(x)=0, expression lim_(x->a)(f(x))/(g(x)) is indeterminate form of type mathbf (0/0).

Case 2. Suppose that lim_(x->a)f(x)=+-oo and lim_(x->a)g(x)=+-oo. We need to find lim_(x->a)(f(x))/(g(x)).

Again it is not very clear what is the limit of the ratio and does it exist at all. Very large number divided by the very large number can give anything.

Let f(x)=1/x^2 and g(x)=1/x^4. Clearly lim_(x->0)1/x^2=oo and lim_(x->0)1/x^4=oo. Then lim_(x->0)(f(x))/(g(x))=lim_(x->0)(1/x^2)/(1/x^4)=lim_(x->0)x^2=0.

Let f(x)=1/x^4 and g(x)=1/x^2. Clearly lim_(x->0)1/x^4=oo and lim_(x->0)1/x^2=oo. Then lim_(x->0)(f(x))/(g(x))=lim_(x->0)(1/x^4)/(1/x^2)=lim_(x->0)1/x^2=oo.

Let f(x)=2/x^2 and g(x)=1/x^2. Clearly lim_(x->0)2/x^2=oo and lim_(x->0)1/x^2=oo. Then lim_(x->0)(f(x))/(g(x))=lim_(x->0)(2/x^2)/(1/x^2)=lim_(x->0)2=2.

Again, in general case we can't find limit of ratio without knowing functions.

To characterize this special case we say that when lim_(x->a)f(x)=+-oo and lim_(x->a)g(x)=+-oo, expression lim_(x->a)(f(x))/(g(x)) is indeterminate form of type mathbf (oo/oo).

Actually we can trasnform this indeterminate form in indeterminate form of type 0/0.

Indeed, since lim_(x->a)f(x)=oo and lim_(x->a)g(x)=oo then lim_(x->a)1/(f(x))=0 and lim_(x->a)1/(g(x))=0.

So, lim_(x->a)(f(x))/(g(x))=lim_(x->a)(1/(g(x)))/(1/(f(x))).

Case 3. Suppose that lim_(x->a)f(x)=0 and lim_(x->a)g(x)=+-oo. We need to find lim_(x->a)f(x)g(x).

Again it is not very clear what is the product of very small number and very large number.

This case can be transformed either into case 1 or into case 2.

Indeed, since lim_(x->a)f(x)=0 then lim_(x->a)1/(f(x))=oo. Therefore, we can write product as indeterminate form of type (oo)/(oo): lim_(x->a)f(x)g(x)=lim_(x->a)(g(x))/(1/(f(x))).

Similarly, since lim_(x->a)g(x)=0 then lim_(x->a)1/(g(x))=0. Therefore, we can write product as indeterminate form of type (0)/(0): lim_(x->a)f(x)g(x)=lim_(x->a)(f(x))/(1/(g(x))).

So, we have third indeterminate form: indeterminate form of type mathbf (0*oo).

Case 4. Finally, suppose that lim_(x->a)f(x)=oo and lim_(x->a)g(x)=oo. We need to find lim_(x->a)(f(x)-g(x)).

Again it is not clear what will be difference of very large numbers.

Let f(x)=x and g(x)=2x. Clearly lim_(x->oo)x=oo and lim_(x->oo)2x=oo. Then lim_(x->oo)(f(x)-g(x))=lim_(x->oo)(x-2x)=lim_(x->oo)-x=-oo.

Let f(x)=3x and g(x)=x. Clearly lim_(x->oo)3x=oo and lim_(x->oo)x=oo. Then lim_(x->oo)(f(x)-g(x))=lim_(x->oo)(3x-x)=lim_(x->oo)2x=oo.

Let f(x)=x+5 and g(x)=x. Clearly lim_(x->oo)(x+5)=oo and lim_(x->oo)x=oo. Then lim_(x->oo)(f(x)-g(x))=lim_(x->oo)(x+5-x)=lim_(x->oo)5=5.

So we can't find limit of difference without knowing functions.

To characterize this special case we say that when lim_(x->a)f(x)=oo and lim_(x->a)g(x)=oo, expression lim_(x->a)((f(x))-(g(x))) is indeterminate form of type mathbf (oo-oo).

So, we have seen four types of indeterminate forms. In these cases we need to know sequences f(x) and g(x). To get rid of indetermination it is often useful to perform algebraic manipulations. Now, let's go through a couple of examples.

Example 1 . Find lim_(x->oo)(3x^2-5x).

Since lim_(x->oo)3x^2=oo and lim_(x->oo)5x=oo we have indeterminate form of type oo-oo.

To handle it, let's perform algebraic manipulations: 3x^2-5x=x^2 (3-5/x).

Now, since lim_(x->oo)x^2=oo and lim_(x->oo)(3-5/x)=3 then their product is also very large number: lim_(x->oo)(3x^2-5x)=oo.

Example 2 . Find lim_(x->oo)(-4x^3+5x^2).

Since lim_(x->oo)-4x^3=-oo and lim_(x->oo)5x^2=oo we have indeterminate form of type oo-oo.

To handle it, let's perform algebraic manipulations: -4x^2+5x^2=x^3 (-4+5/x).

Now, since lim_(x->oo)x^3=oo and lim_(x->oo)(-4+5/x)=-4 then lim_(x->oo)(-4x^3+5x)=-oo.

Example 3 . Find lim_(x->oo)(a_0 x^k+a_1x^(k-1)+...+a_(k-1)x+a_k), where a_0,a_1,...a_k are constants.

This is generalization of above two examples. If all coefficients a_0,a_1,...,a_k have same sign then limit of this sequence is oo (or -oo). But if coeffcients have different signs then we have indeterminate form of type oo-oo.

To handle it, let's perform algebraic manipulations: a_0x^k+a_1x^(k-1)+...+a_(k-1)x+a_k=x^k (a_0+(a_1)/x+...+(a_(k-1))/(x^(k-1))+(a_k)/(x^k)).

Now, since x^k->oo and a_0+(a_1)/x+...+(a_(k-1))/(x^(k-1))+(a_k)/x^k->a_0 as x->oo, then a_0x^k+a_1x^(k-1)+...+a_(k-1)x+a_k->oo if a_0>0 and x_n->-oo if a_0<0.

Example 4 . Find lim_(x->oo)(3x^2-5x)/(7x+3).

Since lim_(x->oo)(3x^2-5x)->oo and lim_(x->oo)(7x+3)->oo we have indeterminate form of type (oo)/(oo).

To handle it, let's perform algebraic manipulations. Factor out x raised to the greatest degree in numerator and denominator (in this case x^2): (3x^2-5x)/(7x+3)=(x^2 (3-5/x))/(x^2(7/x+3/x^2))=(3-5/x^2)/(7/x+3/x^2).

Now, since 3-5/x->3 and 7/x+3/x^2->0 as x->oo, then (3-5/x^2)/(7/x+3/x^2)->oo.

Example 5 . Let lim_(x->oo)(6x^4-3x^2)/(8x^7+3x).

Since lim_(x->oo)(6x^4-3x^2)->oo and lim_(x->oo)(8x^7+3x)->oo we have indeterminate form of type (oo)/(oo).

To handle it, let's perform algebraic manipulations. Factor out x raised to the greatest degree in numerator and denominator (in this case x^7): (6x^4-3x^2)/(8x^7+3x)=(x^7 (6/x^3-3/x^5))/(x^7(8+3/x^6))=(6/x^3-3/x^5)/(8+3/x^6).

Now, since 6/x^3-3/x^5->0 and 8+3/x^6->8 as x->oo, then lim_(x->oo)(6/x^3-3/x^5)/(8+3/x^6)->0.

Example 6. Find lim_(x->oo)(3x^2-5x)/(7x^2+3).

Since lim_(x->oo)(3x^2-5x)=oo and lim_(x->oo)(7x^2+3)=oo we have indeterminate form of type (oo)/(oo).

To handle it, let's perform algebraic manipulations. Factor out x raised to the greatest degree in numerator and denominator (in this case x^2) (3x^2-5x)/(7x^2+3)=(x^2 (3-5/x))/(n^2(7+3/x^2))=(3-5/x)/(7+3/x^2).

Now, since 3-5/x->3 and 7+3/x^2->7 as x->oo, then lim_(x->oo)(3-5/x)/(7+3/x^2)=3/7.

Example 7. Find lim_(x->oo)(a_0x^k+a_1x^(k-1)+...+a_(k-1)x+a_k)/(b_0x^m+b_1x^(m-1)+..+b_(m-1)x+b_m) where a_0,a_1,...,a_k and b_0,b_1,...,b_m are constants.

This is generalization of above three examples. We have indeterminate form of type (oo)/(oo).

To handle it, let's perform algebraic manipulations. Factor out x^k from numerator and x^m from denominator: (a_0x^k+a_1x^(k-1)+...+a_(k-1)x+a_k)/(b_0x^m+b_1x^(m-1)+...+b_(m-1)x+b_m)=(x^k (a_0+(a_1)/x+...+(a_k)/(x^k)))/(x^m(b_0+(b_1)/x+...+(b_m)/(x^m)))=x^(k-m)((a_0+(a_1)/x+...+(a_k)/(x^k))/(b_0+(b_1)/x+...+(b_m)/(x^m))).

Second factor has limit a_0/b_0. If k=m then x^(k-m)=1->1 and required limit approaches (a_0)/(b_0). If k>m then x^(k-m)->oo and required limit approaches oo (or -oo, sign depends on sign of (a_0)/(b_0)). If k<m then x^(k-m)->0 and required limit approaches 0.

Example 8. Find lim_(x->oo)sqrt(x)(sqrt(x+1)-sqrt(x)).

We can rewrite it as lim_(x->oo)(sqrt(x)sqrt(x+1)-sqrt(x)sqrt(x))=lim_(x->oo)(sqrt(x^2+x)-x).

This indeterminate form of type oo-oo.

When we deal with radicals, we need to multiply both numerator and denominator by conjugate radical. Here we multiply both numerator and denominator by sqrt(x^2+x)+x:

sqrt(x^2+x)-x=((sqrt(x^2+x)-x)color(red)((sqrt(x^2+x)+x)))/(color(red)(sqrt(x^2+x)+sqrt(x)))=((sqrt(x^2+x))^2-x^2)/(sqrt(x^2+x)+sqrt(x))=(x^2+x-x^2)/(sqrt(x^2+x)+x)=

=(x)/(sqrt(x^2+x)+x).

Now, factor out x: x/(sqrt(x^2+x)+x)=(x/(x(sqrt(1+1/x)+1)))=1/(sqrt(1+1/x)-1).

Since 1<sqrt(1+1/x)<1+1/x for large x and 1+1/x->1 as x->oo then by Squeeze Theorem sqrt(1+1/x)->1.

That's why lim_(x->oo)(sqrt(x^2+x)-x)=lim_(x->oo)(1/(sqrt(1+1/x)-1))=1/(1+1)=1/2.

Now, we will calculate limit that we will need later.

Example 9. Find lim_(x->0)(sin(x))/x.

As x->0 sin(x)->0, so we have indeterminate form of type 0/0.

Before finding this limit we need to prove that for 0<x<pi/2 we have that sin(x)<x<tan(x).

For this consider circle with radius r, tangent AC to circle and central angle x.

Clearly area of triangle AOB is less than area of sector AOB, and area of sector AOB is less than area of triangle AOC.

We have that area of sector AOB is (x)/(2pi)*pir^2=1/2 r^2x.

Since area of triangle AOB is 1/2 r^2sin(x), area of sector AOB is 1/2r^2x and area of triangle AOC is 1/2 r^2tan(x) then 1/2r^2sin(x)<1/2r^2x<1/2r^2tan(x) or sin(x)<x<tan(x).

Now, since 0<x<pi/2 then we can divide sin(x) by each member of inequality (don't forget to invert inequality signs): (sin(x))/(color(red)(sin(x)))>(sin(x))/(color(red)(x))>(sin(x))/(color(red)(tan(x))) or 1>(sin(x))/x>cos(x).

Thus, we obtained that cos(x)<(sin(x))/x<1.

As x->0 cos(x)->1 that's why lim_(x->0)cos(x)=1, so, by Squeeze Theorem lim_(x->0^+)(sin(x))/x=1. (we wrote one-sided limit because we considered interval (0,pi/2)). Since (sin(x))/x is even function then also lim_(x->0^-)(sin(x))/x=1. Since one-sided limits are equal then lim_(x->0)(sin(x))/x=1.

From this limit we can find other limits.

Example 10. Find lim_(x->0)(1-cos(x))/x^2.

Here we have indeterminate form of type 0/0.

Since 1-cos(x)=2sin^2(x/2) then (1-cos(x))/x^2=(2sin^2(x/2))/x^2=1/2 (sin^2(x/2))/((x^2)/4)=1/2 ((sin(x/2))/(x/2))^2.

Since x->0 then x/2->0 and so lim_(x->0)(sin(x/2))/(x/2)=1.

Thus, lim_(x->0)(1-cos(x))/x^2=lim_(x->0)1/2 ((sin(x/2))/(x/2))^2=1/2 (lim_(x->0)(sin(x/2))/(x/2))^2=1/2*1^2=1/2.

When we talked about number color(254,126,0)(e) we proved that sequence x_n=(1+1/n)^n has limit e.

We have same thing for functions and formulate it as following fact:

Fact. lim_(x->oo)(1+1/x)^x=e, lim_(x->-oo)(1+1/x)^x=e.

If we make substitution t=1/x then t->0 as x->oo. Thus, lim_(t->0)(1+t)^(1/t)=e.

Example 11. Prove that lim_(h->0)(a^h-1)/h=ln(a).

Let a^h-1=x then x->0 as h->0 . And h=log_a(x+1)=ln(x+1)/ln(a)

So, lim_(h->0)(a^h-1)/h=lim_(x->0)x/((ln(x+1))/(ln(a)))=ln(a) lim_(x->0)x/(ln(x+1))=ln(a) lim_(x->0) 1/(1/x ln(x+1))=

=ln(a) lim_(x->0) 1/(ln(1+x)^(1/x))=ln(a) 1/(ln(lim_(x->0)(1+x)^(1/x)))=ln(a) 1/(ln(e))=ln(a) 1/1=ln(a)

Now, suppose that lim_(x->a)f(x)=L and lim_(x->a)g(x)=M. We want to find lim_(x->a)f(x)^(g(x)).

To handle it we use continuity of the natural logarithm: let lim_(x->a)f(x)^(g(x))=B then ln(lim_(x->a)f(x)^(g(x)))=ln(B).

This can be rewritten as lim_(x->a)ln(f(x)^(g(x)))=ln(B) or lim_(x->a)(g(x)ln(f(x)))=ln(B).

Again using continuity of the function we can write that lim_(x->a)g(x)*ln(lim_(x->a)f(x))=ln(B).

So, ln(B)=Mln(L) or B=L^M.

We can always find this limit except three indeterminate cases:

1. L=1, M=+-oo. In this case Mln(L) is indeterminate form of type oo*0.
2. L=0, M=0. In this case Mln(L) is indeterminate form of type 0*oo.
3. L=+oo, M=0. In this case Mln(L) is indeterminate form of type 0*oo.

Example 12. Find lim_(x->0)x^x.

Since lim_(x->0)x=0 then we have case b) here.

Let B=x^x then ln(B)=ln(x^x)=xln(x).

It is known that lim_(x->0)xln(x)=0 that's why ln(B)=0 or B=1.

Thus, lim_(x->0)x^x=1.