# Indeterminate Form for Sequence

## Related Calculator: Limit Calculator

When we described arithmetic operations on limits, we made assumption that sequences approach finite limits.

Now, let's consider case when limits are infinite or, in the case of quotient, limit of denominator equals 0.

There are four basic cases.

Case 1. Consider quotient (x_n)/(y_n) where x_n->0 and y_n->0. Here we first face with special situation: although we know limit of sequences x_n and y_n, but we can't say what is limit of ratio (x_n)/(y_n) without knowing sequences x_n and y_n. Limit of ratio can have different values or even doesn't exist.

Let x_n=1/n^2 and y_n=1/n. Clearly x_n->0 and y_n->0. Then (x_n)/(y_n)=(1/n^2)/(1/n)=1/n->0.

Now let x_n=1/n and y_n=1/n^2. Clearly x_n->0 and y_n->0. Then (x_n)/(y_n)=(1/n)/(1/n^2)=n->oo.

Now let's take number a!=0 and construct sequences x_n=a/n and y_n=1/n. Clearly x_n->0 and y_n->0. Then (x_n)/(y_n)=(a/n)/(1/n)=a->a (because sequence is constant and each member equals a).

At last let x_n=((-1)^(n+1))/n and y_n=1/n (both have limit 0), but (x_n)/(y_n)=(-1)^(n+1) doesn't have limit.

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

To characterize this special case we say that when x_n->0 and y_n->0, expression (x_n)/(y_n) is indeterminate form of type mathbf (0/0).

Case 2. When simultaneously x_n->+-oo and y_n->+-oo, we face with same situation. Without knowing sequences we can't say what is limit of their ratio.

Let x_n=n and y_n=n^2. Clearly x_n->oo and y_n->oo. Then (x_n)/(y_n)=1/n->0.

Now let x_n=n^2 and y_n=n. Clearly x_n->oo and y_n->oo. Then (x_n)/(y_n)=n->oo.

Now let's take number a>0 and construct sequences x_n=an and y_n=n. Clearly x_n->oo and y_n->oo. Then (x_n)/(y_n)=a->a (because sequence is constant and each member equals a).

At last let x_n=(2+(-1)^(n+1))n and y_n=n (both have limit oo), but (x_n)/(y_n)=2+(-1)^(n+1) doesn't have limit.

In this case we say that when x_n->+-oo and y_n->+-oo, expression (x_n)/(y_n) is indeterminate form of type mathbf ((oo)/(oo)).

Case 3. Now consider product x_n y_n. Here we will have indetermination when x_n->0 and y_n->+-oo or vice versa.

If we take b_n=1/y_n then b_n->0. So, x_n y_n=(x_n)/(1/y_n)=(x_n)/(b_n) and we again have indeterminate form of type 0/0.

Therefore, when x_n->0 and y_n->+-oo we say that expression x_n y_n is indeterminate form of type mathbf (0*oo). This form can be transformed into indeterminate form of type 0/0.

Case 4. Finally consider sum x_n+y_n. Here we will have indeterminate form when x_n and y_n approach infinity with with diffferent signs.

Let x_n=2n and y_n=-n. Clearly x_n->oo and y_n->-oo. Then x_n+y_n=2n-n=n>oo.

Now let x_n=n and y_n=-2n. Clearly x_n->oo and y_n-oo. Then x_n+y_n=n-2n=-n->-oo.

Let x_n=a+n and y_n=-n. Clearly x_n->oo and y_n->-oo. Then x_n+y_n=a+n-a=a->a (because sequence is constant and each member equals a).

At last let x_n=n+(-1)^(n+1) and y_n=-n. In this case x_n+y_n=(-1)^(n+1) doesn't have limit.

In this case we say that when x_n->+oo and y_n->-oo, expression x_n+y_n 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 x_n and y_n. To get rid of indetermination it is often useful to perform algebraic manipulations. Now, let's go through a couple of examples.

Example 1 . Let x_n=3n^2-5n. Find limit of this sequence.

Since 3n^2->oo and -5n->-oo we have indeterminate form of type oo-oo.

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

Now, since n^2->oo and 3-5/n->3 then x_n->oo.

Example 2 . Let x_n=-4n^3+5n^2. Find limit of this sequence.

Since -4n^3->-oo and 5n^2->oo we have indeterminate form of type oo-oo.

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

Now, since n^3->oo and -4+5/n->-4 then x_n->-oo.

Example 3 . Let x_n=a_0 n^k+a_1n^(k-1)+...+a_(k-1)n+a_k, where a_0,a_1,...a_k are constants. Find limit of this sequence.

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: x_n=n^k (a_0+(a_1)/n+...+(a_(k-1))/(n^(k-1))+(a_k)/(n^k)).

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

Example 4. Let x_n=(3n^2-5n)/(7n+3). Find limit of this sequence.

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

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

Now, since 3-5/n->3 and 7/n+3/n^2->0 then x_n->oo.

Example 5. Let x_n=(6n^4-3n^2)/(8n^7+3). Find limit of this sequence.

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

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

Now, since 6/n^3-3/n^5->0 and 8+3/n^6->8 then x_n->0.

Example 6. Let x_n=(3n^2-5n)/(7n^2+3). Find limit of this sequence.

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

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

Now, since 3-5/n->3 and 7+3/n^2->7 then x_n->3/7.

Example 7. Let x_n=(a_0n^k+a_1n^(k-1)+...+a_(k-1)n+a_k)/(b_0n^m+b_1n^(m-1)+..+b_(m-1)n+b_m) where a_0,a_1,...,a_k and b_0,b_1,...,b_m are constants. Find limit of this sequence.

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 n^k from numerator and n^m from denominator: x_n=(n^k (a_0+(a_1)/n+...+(a_k)/(n^k)))/(n^m(b_0+(b_1)/n+...+(b_m)/(n^m)))=n^(k-m)((a_0+(a_1)/n+...+(a_k)/(n^k))/(b_0+(b_1)/n+...+(b_m)/(n^m))).

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

Example 8. Prove that for 0<k<1 lim((n+1)^k-n^k)=0.

We have indeterminate form of type oo-oo.

We have that 0<(n+1)^k-n^k=n^k((1+1/k)^k-1)<n^k((1+1/n)-1)=n^(k-1)=1/(n^(1-k)).

So, we obtained that 0<(n+1)^k-n^k<1/(n^(1-k)).

Since 1/(n^(1-k))->0 then by Squeeze Theorem (n+1)^k-n^k->0.

Example 9. Find limit of x_n=sqrt(n)(sqrt(n+1)-sqrt(n)).

According to example 8 sqrt(n+1)-sqrt(n)->0 that's why we have intedeterminate form of type oo*0.

Let's transform this indeterminate form into indeterminate form of type (oo)/(oo). To do this multiply both numerator and denominator by sqrt(n+1)+sqrt(n):

x_n=(sqrt(n)(sqrt(n+1)-sqrt(n))color(red)((sqrt(n+1)+sqrt(n+1))))/(color(red)(sqrt(n+1)+sqrt(n)))=(sqrt(n)((sqrt(n+1))^2-(sqrt(n))^2))/(sqrt(n+1)+sqrt(n))=(sqrt(n)(n+1-n))/(sqrt(n+1)+sqrt(n))=

=(sqrt(n))/(sqrt(n+1)+sqrt(n)).

Now, factor out sqrt(n): x_n=(sqrt(n))/(sqrt(n)(sqrt(1+1/n)+1))=1/(sqrt(1+1/n)+1).

Since 1<sqrt(1+1/n)<1+1/n and 1+1/n->1 then by squeeze theorem sqrt(1+1/n)->1.

That's why x_n=1/(sqrt(1+1/n)-1)->1/(1+1)=1/2.

Example 10. Find limits of x_n=n/(sqrt(n^2+n)), y_n=n/(sqrt(n^2+1)) and z_n=1/(sqrt(n^2+1))+1/(sqrt(n^2+2))+...+1/(sqrt(n^2+n)).

Sequences x_n and y_n are indeterminate forms of type (oo)/(oo).

Since x_n=n/(sqrt(n^2+n))=n/(n sqrt(1+1/n))=1/(sqrt(1+1/n)) and sqrt(1+1/n)->1 then x_n->1.

Also, since y_n=n/(sqrt(n^2+1))=n/(n sqrt(1+1/n^2))=1/(sqrt(1+1/n^2)) and sqrt(1+1/n^2)->1 then y_n->1.

Now let's find limit of z_n. Formula for z_n contains n summands and each summand is less than previous, therefore,

1/sqrt(n^2+n)+1/sqrt(n^2+n)+...+1/sqrt(n^2+n)<z_n<1/sqrt(n^2+1)+1/sqrt(n^2+1)+...+1/sqrt(n^2+1) or

n/sqrt(n^2+n)<z_n<n/sqrt(n^2+1) i.e. x_n<z_n<y_n.

Since x_n->1 and y_n->1 then by Squeeze Theorem z_n->1.

Example 11. Suppose we are given m positive numbers a_1,a_2,...,a_m. Find limit of x_n=root(n)(a_1^n+a_2^n+...+a_m^n).

If we denote the greatest of m positive integers by A then

root(n)(A^n+0+...+0)<root(n)(a_1^n+a_2^n+...+a_m^n)<root(n)(A+A+...+A) or A<root(n)(a_1^n+a_2^n+...+a_m^n)<Aroot(n)(m).

Since root(n)(m)->1 then by Squeeze Theorem x_n=root(n)(a_1^n+a_2^n+...+a_m^n)->A.