# 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 a quotient $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}$ where ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to{0}$. 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 $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}$ without knowing sequences ${x}_{{n}}$ and ${y}_{{n}}$. Limit of ratio can have different values or even doesn't exist.

Let ${x}_{{n}}=\frac{{1}}{{{n}}^{{2}}}$ and ${y}_{{n}}=\frac{{1}}{{n}}$. Clearly ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to{0}$. Then $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{\frac{{1}}{{{n}}^{{2}}}}}{{\frac{{1}}{{n}}}}=\frac{{1}}{{n}}\to{0}$.

Now let ${x}_{{n}}=\frac{{1}}{{n}}$ and ${y}_{{n}}=\frac{{1}}{{{n}}^{{2}}}$. Clearly ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to{0}$. Then $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{\frac{{1}}{{n}}}}{{\frac{{1}}{{{n}}^{{2}}}}}={n}\to\infty$.

Now let's take number ${a}\ne{0}$ and construct sequences ${x}_{{n}}=\frac{{a}}{{n}}$ and ${y}_{{n}}=\frac{{1}}{{n}}$. Clearly ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to{0}$. Then $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{\frac{{a}}{{n}}}}{{\frac{{1}}{{n}}}}={a}\to{a}$ (because sequence is constant and each member equals ${a}$).

At last let ${x}_{{n}}=\frac{{{{\left(-{1}\right)}}^{{{n}+{1}}}}}{{n}}$ and ${y}_{{n}}=\frac{{1}}{{n}}$ (both have limit 0), but $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}={{\left(-{1}\right)}}^{{{n}+{1}}}$ 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 ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to{0}$, expression $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}$ is indeterminate form of type $\frac{0}{0}$.

Case 2. When simultaneously ${x}_{{n}}\to\pm\infty$ and ${y}_{{n}}\to\pm\infty$, 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 ${x}_{{n}}\to\infty$ and ${y}_{{n}}\to\infty$. Then $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{1}}{{n}}\to{0}$.

Now let ${x}_{{n}}={{n}}^{{2}}$ and ${y}_{{n}}={n}$. Clearly ${x}_{{n}}\to\infty$ and ${y}_{{n}}\to\infty$. Then $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}={n}\to\infty$.

Now let's take number ${a}>{0}$ and construct sequences ${x}_{{n}}={a}{n}$ and ${y}_{{n}}={n}$. Clearly ${x}_{{n}}\to\infty$ and ${y}_{{n}}\to\infty$. Then $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}={a}\to{a}$ (because sequence is constant and each member equals ${a}$).

At last let ${x}_{{n}}={\left({2}+{{\left(-{1}\right)}}^{{{n}+{1}}}\right)}{n}$ and ${y}_{{n}}={n}$ (both have limit $\infty$), but $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}={2}+{{\left(-{1}\right)}}^{{{n}+{1}}}$ doesn't have limit.

In this case we say that when ${x}_{{n}}\to\pm\infty$ and ${y}_{{n}}\to\pm\infty$, expression $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}$ is indeterminate form of type $\frac{\infty}{\infty}$.

Case 3. Now consider product ${x}_{{n}}{y}_{{n}}$. Here we will have indetermination when ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to\pm\infty$ or vice versa.

If we take ${b}_{{n}}=\frac{{1}}{{y}_{{n}}}$ then ${b}_{{n}}\to{0}$. So, ${x}_{{n}}{y}_{{n}}=\frac{{{x}_{{n}}}}{{\frac{{1}}{{y}_{{n}}}}}=\frac{{{x}_{{n}}}}{{{b}_{{n}}}}$ and we again have indeterminate form of type $\frac{{0}}{{0}}$.

Therefore, when ${x}_{{n}}\to{0}$ and ${y}_{{n}}\to\pm\infty$ we say that expression ${x}_{{n}}{y}_{{n}}$ is indeterminate form of type $0\cdot\infty$. This form can be transformed into indeterminate form of type $\frac{{0}}{{0}}$.

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 ${x}_{{n}}={2}{n}$ and ${y}_{{n}}=-{n}$. Clearly ${x}_{{n}}\to\infty$ and ${y}_{{n}}\to-\infty$. Then ${x}_{{n}}+{y}_{{n}}={2}{n}-{n}={n}>\infty$.

Now let ${x}_{{n}}={n}$ and ${y}_{{n}}=-{2}{n}$. Clearly ${x}_{{n}}\to\infty$ and ${y}_{{n}}-\infty$. Then ${x}_{{n}}+{y}_{{n}}={n}-{2}{n}=-{n}\to-\infty$.

Let ${x}_{{n}}={a}+{n}$ and ${y}_{{n}}=-{n}$. Clearly ${x}_{{n}}\to\infty$ and ${y}_{{n}}\to-\infty$. Then ${x}_{{n}}+{y}_{{n}}={a}+{n}-{a}={a}\to{a}$ (because sequence is constant and each member equals ${a}$).

At last let ${x}_{{n}}={n}+{{\left(-{1}\right)}}^{{{n}+{1}}}$ and ${y}_{{n}}=-{n}$. In this case ${x}_{{n}}+{y}_{{n}}={{\left(-{1}\right)}}^{{{n}+{1}}}$ doesn't have limit.

In this case we say that when ${x}_{{n}}\to+\infty$ and ${y}_{{n}}\to-\infty$, expression ${x}_{{n}}+{y}_{{n}}$ is indeterminate form of type $\infty-\infty$.

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 ${x}_{{n}}={3}{{n}}^{{2}}-{5}{n}$. Find limit of this sequence.

Since ${3}{{n}}^{{2}}\to\infty$ and $-{5}{n}\to-\infty$ we have indeterminate form of type $\infty-\infty$.

To handle it, let's perform algebraic manipulations: ${x}_{{n}}={{n}}^{{2}}{\left({3}-\frac{{5}}{{n}}\right)}$.

Now, since ${{n}}^{{2}}\to\infty$ and ${3}-\frac{{5}}{{n}}\to{3}$ then ${x}_{{n}}\to\infty$.

Example 2. Let ${x}_{{n}}=-{4}{{n}}^{{3}}+{5}{{n}}^{{2}}$. Find limit of this sequence.

Since $-{4}{{n}}^{{3}}\to-\infty$ and ${5}{{n}}^{{2}}\to\infty$ we have indeterminate form of type $\infty-\infty$.

To handle it, let's perform algebraic manipulations: ${x}_{{n}}={{n}}^{{3}}{\left(-{4}+\frac{{5}}{{n}}\right)}$.

Now, since ${{n}}^{{3}}\to\infty$ and $-{4}+\frac{{5}}{{n}}\to-{4}$ then ${x}_{{n}}\to-\infty$.

Example 3. Let ${x}_{{n}}={a}_{{0}}{{n}}^{{k}}+{a}_{{1}}{{n}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{n}+{a}_{{k}}$, where ${a}_{{0}},{a}_{{1}},\ldots{a}_{{k}}$ are constants. Find limit of this sequence.

This is generalization of above two examples. If all coefficients ${a}_{{0}},{a}_{{1}},\ldots,{a}_{{k}}$ have same sign then limit of this sequence is $\infty$ (or $-\infty$). But if coeffcients have different signs then we have indeterminate form of type $\infty-\infty$.

To handle it, let's perform algebraic manipulations: ${x}_{{n}}={{n}}^{{k}}{\left({a}_{{0}}+\frac{{{a}_{{1}}}}{{n}}+\ldots+\frac{{{a}_{{{k}-{1}}}}}{{{{n}}^{{{k}-{1}}}}}+\frac{{{a}_{{k}}}}{{{{n}}^{{k}}}}\right)}$.

Now, since ${{n}}^{{k}}\to\infty$ and ${a}_{{0}}+\frac{{{a}_{{1}}}}{{n}}+\ldots+\frac{{{a}_{{{k}-{1}}}}}{{{{n}}^{{{k}-{1}}}}}+\frac{{{a}_{{k}}}}{{{{n}}^{{k}}}}\to{a}_{{0}}$ then ${x}_{{n}}\to\infty$ if ${a}_{{0}}>{0}$ and ${x}_{{n}}\to-\infty$ if ${a}_{{0}}<{0}$.

Example 4. Let ${x}_{{n}}=\frac{{{3}{{n}}^{{2}}-{5}{n}}}{{{7}{n}+{3}}}$. Find limit of this sequence.

Since ${3}{{n}}^{{2}}-{5}{n}\to\infty$ and ${7}{n}+{3}\to\infty$ we have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${n}$ raised to the greatest degree in numerator and denominator (in this case ${{n}}^{{2}}$): ${x}_{{n}}=\frac{{{{n}}^{{2}}{\left({3}-\frac{{5}}{{n}}\right)}}}{{{{n}}^{{2}}{\left(\frac{{7}}{{n}}+\frac{{3}}{{{n}}^{{2}}}\right)}}}=\frac{{{3}-\frac{{5}}{{n}}}}{{\frac{{7}}{{n}}+\frac{{3}}{{{n}}^{{2}}}}}$.

Now, since ${3}-\frac{{5}}{{n}}\to{3}$ and $\frac{{7}}{{n}}+\frac{{3}}{{{n}}^{{2}}}\to{0}$ then ${x}_{{n}}\to\infty$.

Example 5. Let ${x}_{{n}}=\frac{{{6}{{n}}^{{4}}-{3}{{n}}^{{2}}}}{{{8}{{n}}^{{7}}+{3}}}$. Find limit of this sequence.

Since ${6}{{n}}^{{4}}-{3}{{n}}^{{2}}\to\infty$ and ${8}{{n}}^{{7}}+{3}{n}\to\infty$ we have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${n}$ raised to the greatest degree in numerator and denominator (in this case ${{n}}^{{7}}$): ${x}_{{n}}=\frac{{{{n}}^{{7}}{\left(\frac{{6}}{{{n}}^{{3}}}-\frac{{3}}{{{n}}^{{5}}}\right)}}}{{{{n}}^{{7}}{\left({8}+\frac{{3}}{{{n}}^{{6}}}\right)}}}=\frac{{\frac{{6}}{{{n}}^{{3}}}-\frac{{3}}{{{n}}^{{5}}}}}{{{8}+\frac{{3}}{{{n}}^{{6}}}}}$.

Now, since $\frac{{6}}{{{n}}^{{3}}}-\frac{{3}}{{{n}}^{{5}}}\to{0}$ and ${8}+\frac{{3}}{{{n}}^{{6}}}\to{8}$ then ${x}_{{n}}\to{0}$.

Example 6. Let ${x}_{{n}}=\frac{{{3}{{n}}^{{2}}-{5}{n}}}{{{7}{{n}}^{{2}}+{3}}}$. Find limit of this sequence.

Since ${3}{{n}}^{{2}}-{5}{n}\to\infty$ and ${7}{{n}}^{{2}}+{3}\to\infty$ we have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${n}$ raised to the greatest degree in numerator and denominator (in this case ${{n}}^{{2}}$): ${x}_{{n}}=\frac{{{{n}}^{{2}}{\left({3}-\frac{{5}}{{n}}\right)}}}{{{{n}}^{{2}}{\left({7}+\frac{{3}}{{{n}}^{{2}}}\right)}}}=\frac{{{3}-\frac{{5}}{{n}}}}{{{7}+\frac{{3}}{{{n}}^{{2}}}}}$.

Now, since ${3}-\frac{{5}}{{n}}\to{3}$ and ${7}+\frac{{3}}{{{n}}^{{2}}}\to{7}$ then ${x}_{{n}}\to\frac{{3}}{{7}}$.

Example 7. Let ${x}_{{n}}=\frac{{{a}_{{0}}{{n}}^{{k}}+{a}_{{1}}{{n}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{n}+{a}_{{k}}}}{{{b}_{{0}}{{n}}^{{m}}+{b}_{{1}}{{n}}^{{{m}-{1}}}+..+{b}_{{{m}-{1}}}{n}+{b}_{{m}}}}$ where ${a}_{{0}},{a}_{{1}},\ldots,{a}_{{k}}$ and ${b}_{{0}},{b}_{{1}},\ldots,{b}_{{m}}$ are constants. Find limit of this sequence.

This is generalization of above three examples. We have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${{n}}^{{k}}$ from numerator and ${{n}}^{{m}}$ from denominator: ${x}_{{n}}=\frac{{{{n}}^{{k}}{\left({a}_{{0}}+\frac{{{a}_{{1}}}}{{n}}+\ldots+\frac{{{a}_{{k}}}}{{{{n}}^{{k}}}}\right)}}}{{{{n}}^{{m}}{\left({b}_{{0}}+\frac{{{b}_{{1}}}}{{n}}+\ldots+\frac{{{b}_{{m}}}}{{{{n}}^{{m}}}}\right)}}}={{n}}^{{{k}-{m}}}{\left(\frac{{{a}_{{0}}+\frac{{{a}_{{1}}}}{{n}}+\ldots+\frac{{{a}_{{k}}}}{{{{n}}^{{k}}}}}}{{{b}_{{0}}+\frac{{{b}_{{1}}}}{{n}}+\ldots+\frac{{{b}_{{m}}}}{{{{n}}^{{m}}}}}}\right)}$.

Second factor has limit $\frac{{a}_{{0}}}{{b}_{{0}}}$. If ${k}={m}$ then ${{n}}^{{{k}-{m}}}={1}\to{1}$ and ${x}_{{n}}\to\frac{{{a}_{{0}}}}{{{b}_{{0}}}}$. If ${k}>{m}$ then ${{n}}^{{{k}-{m}}}\to\infty$ and ${x}_{{n}}\to\infty$ (or $-\infty$, sign depends on sign of $\frac{{{a}_{{0}}}}{{{b}_{{0}}}}$). If ${k}<{m}$ then ${{n}}^{{{k}-{m}}}\to{0}$ and ${x}_{{n}}\to{0}$.

Example 8. Prove that for ${0}<{k}<{1}$ $\lim{\left({{\left({n}+{1}\right)}}^{{k}}-{{n}}^{{k}}\right)}={0}$.

We have indeterminate form of type $\infty-\infty$.

We have that ${0}<{{\left({n}+{1}\right)}}^{{k}}-{{n}}^{{k}}={{n}}^{{k}}{\left({{\left({1}+\frac{{1}}{{k}}\right)}}^{{k}}-{1}\right)}<{{n}}^{{k}}{\left({\left({1}+\frac{{1}}{{n}}\right)}-{1}\right)}={{n}}^{{{k}-{1}}}=\frac{{1}}{{{{n}}^{{{1}-{k}}}}}$.

So, we obtained that ${0}<{{\left({n}+{1}\right)}}^{{k}}-{{n}}^{{k}}<\frac{{1}}{{{{n}}^{{{1}-{k}}}}}$.

Since $\frac{{1}}{{{{n}}^{{{1}-{k}}}}}\to{0}$ then by Squeeze Theorem ${{\left({n}+{1}\right)}}^{{k}}-{{n}}^{{k}}\to{0}$.

Example 9. Find limit of ${x}_{{n}}=\sqrt{{{n}}}{\left(\sqrt{{{n}+{1}}}-\sqrt{{{n}}}\right)}$.

According to example 8 $\sqrt{{{n}+{1}}}-\sqrt{{{n}}}\to{0}$ that's why we have intedeterminate form of type $\infty\cdot{0}$.

Let's transform this indeterminate form into indeterminate form of type $\frac{{\infty}}{{\infty}}$. To do this multiply both numerator and denominator by $\sqrt{{{n}+{1}}}+\sqrt{{{n}}}$:

${x}_{{n}}=\frac{{\sqrt{{{n}}}{\left(\sqrt{{{n}+{1}}}-\sqrt{{{n}}}\right)}{\color{red}{{{\left(\sqrt{{{n}+{1}}}+\sqrt{{{n}+{1}}}\right)}}}}}}{{{\color{red}{{\sqrt{{{n}+{1}}}+\sqrt{{{n}}}}}}}}=\frac{{\sqrt{{{n}}}{\left({{\left(\sqrt{{{n}+{1}}}\right)}}^{{2}}-{{\left(\sqrt{{{n}}}\right)}}^{{2}}\right)}}}{{\sqrt{{{n}+{1}}}+\sqrt{{{n}}}}}=\frac{{\sqrt{{{n}}}{\left({n}+{1}-{n}\right)}}}{{\sqrt{{{n}+{1}}}+\sqrt{{{n}}}}}=$

$=\frac{{\sqrt{{{n}}}}}{{\sqrt{{{n}+{1}}}+\sqrt{{{n}}}}}$.

Now, factor out $\sqrt{{{n}}}$: ${x}_{{n}}=\frac{{\sqrt{{{n}}}}}{{\sqrt{{{n}}}{\left(\sqrt{{{1}+\frac{{1}}{{n}}}}+{1}\right)}}}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{n}}}}+{1}}}$.

Since ${1}<\sqrt{{{1}+\frac{{1}}{{n}}}}<{1}+\frac{{1}}{{n}}$ and ${1}+\frac{{1}}{{n}}\to{1}$ then by squeeze theorem $\sqrt{{{1}+\frac{{1}}{{n}}}}\to{1}$.

That's why ${x}_{{n}}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{n}}}}-{1}}}\to\frac{{1}}{{{1}+{1}}}=\frac{{1}}{{2}}$.

Example 10. Find limits of ${x}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{n}}}}}$, ${y}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{1}}}}}$ and ${z}_{{n}}=\frac{{1}}{{\sqrt{{{{n}}^{{2}}+{1}}}}}+\frac{{1}}{{\sqrt{{{{n}}^{{2}}+{2}}}}}+\ldots+\frac{{1}}{{\sqrt{{{{n}}^{{2}}+{n}}}}}$.

Sequences ${x}_{{n}}$ and ${y}_{{n}}$ are indeterminate forms of type $\frac{{\infty}}{{\infty}}$.

Since ${x}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{n}}}}}=\frac{{n}}{{{n}\sqrt{{{1}+\frac{{1}}{{n}}}}}}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{n}}}}}}$ and $\sqrt{{{1}+\frac{{1}}{{n}}}}\to{1}$ then ${x}_{{n}}\to{1}$.

Also, since ${y}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{1}}}}}=\frac{{n}}{{{n}\sqrt{{{1}+\frac{{1}}{{{n}}^{{2}}}}}}}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{{n}}^{{2}}}}}}}$ and $\sqrt{{{1}+\frac{{1}}{{{n}}^{{2}}}}}\to{1}$ then ${y}_{{n}}\to{1}$.

Now let's find limit of ${z}_{{n}}$. Formula for ${z}_{{n}}$ contains ${n}$ summands and each summand is less than previous, therefore,

$\frac{{1}}{\sqrt{{{{n}}^{{2}}+{n}}}}+\frac{{1}}{\sqrt{{{{n}}^{{2}}+{n}}}}+\ldots+\frac{{1}}{\sqrt{{{{n}}^{{2}}+{n}}}}<{z}_{{n}}<\frac{{1}}{\sqrt{{{{n}}^{{2}}+{1}}}}+\frac{{1}}{\sqrt{{{{n}}^{{2}}+{1}}}}+\ldots+\frac{{1}}{\sqrt{{{{n}}^{{2}}+{1}}}}$ or

$\frac{{n}}{\sqrt{{{{n}}^{{2}}+{n}}}}<{z}_{{n}}<\frac{{n}}{\sqrt{{{{n}}^{{2}}+{1}}}}$ i.e. ${x}_{{n}}<{z}_{{n}}<{y}_{{n}}$.

Since ${x}_{{n}}\to{1}$ and ${y}_{{n}}\to{1}$ then by Squeeze Theorem ${z}_{{n}}\to{1}$.

Example 11. Suppose we are given ${m}$ positive numbers ${a}_{{1}},{a}_{{2}},\ldots,{a}_{{m}}$. Find limit of ${x}_{{n}}={\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}$.

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

${\sqrt[{{n}}]{{{{A}}^{{n}}+{0}+\ldots+{0}}}}<{\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}<{\sqrt[{{n}}]{{{A}+{A}+\ldots+{A}}}}$ or ${A}<{\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}<{A}{\sqrt[{{n}}]{{{m}}}}$.

Since ${\sqrt[{{n}}]{{{m}}}}\to{1}$ then by Squeeze Theorem ${x}_{{n}}={\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}\to{A}$.