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Indeterminate Form for Sequence

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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 xnyn\frac{{{x}_{{n}}}}{{{y}_{{n}}}} where xn→0{x}_{{n}}\to{0} and yn→0{y}_{{n}}\to{0}. Here we first face with special situation: although we know limit of sequences xn{x}_{{n}} and yn{y}_{{n}}, but we can't say what is limit of ratio xnyn\frac{{{x}_{{n}}}}{{{y}_{{n}}}} without knowing sequences xn{x}_{{n}} and yn{y}_{{n}}. Limit of ratio can have different values or even doesn't exist.

Let xn=1n2{x}_{{n}}=\frac{{1}}{{{n}}^{{2}}} and yn=1n{y}_{{n}}=\frac{{1}}{{n}}. Clearly xn→0{x}_{{n}}\to{0} and yn→0{y}_{{n}}\to{0}. Then xnyn=1n21n=1n→0\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{\frac{{1}}{{{n}}^{{2}}}}}{{\frac{{1}}{{n}}}}=\frac{{1}}{{n}}\to{0}.

Now let xn=1n{x}_{{n}}=\frac{{1}}{{n}} and yn=1n2{y}_{{n}}=\frac{{1}}{{{n}}^{{2}}}. Clearly xn→0{x}_{{n}}\to{0} and yn→0{y}_{{n}}\to{0}. Then xnyn=1n1n2=n→∞\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{\frac{{1}}{{n}}}}{{\frac{{1}}{{{n}}^{{2}}}}}={n}\to\infty.

Now let's take number a≠0{a}\ne{0} and construct sequences xn=an{x}_{{n}}=\frac{{a}}{{n}} and yn=1n{y}_{{n}}=\frac{{1}}{{n}}. Clearly xn→0{x}_{{n}}\to{0} and yn→0{y}_{{n}}\to{0}. Then xnyn=an1n=a→a\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{\frac{{a}}{{n}}}}{{\frac{{1}}{{n}}}}={a}\to{a} (because sequence is constant and each member equals a{a}).

At last let xn=(−1)n+1n{x}_{{n}}=\frac{{{{\left(-{1}\right)}}^{{{n}+{1}}}}}{{n}} and yn=1n{y}_{{n}}=\frac{{1}}{{n}} (both have limit 0), but xnyn=(−1)n+1\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 xn→0{x}_{{n}}\to{0} and yn→0{y}_{{n}}\to{0}, expression xnyn\frac{{{x}_{{n}}}}{{{y}_{{n}}}} is indeterminate form of type 00\frac{0}{0}.

Case 2. When simultaneously xn→±∞{x}_{{n}}\to\pm\infty and yn→±∞{y}_{{n}}\to\pm\infty, we face with same situation. Without knowing sequences we can't say what is limit of their ratio.

Let xn=n{x}_{{n}}={n} and yn=n2{y}_{{n}}={{n}}^{{2}}. Clearly xn→∞{x}_{{n}}\to\infty and yn→∞{y}_{{n}}\to\infty. Then xnyn=1n→0\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{1}}{{n}}\to{0}.

Now let xn=n2{x}_{{n}}={{n}}^{{2}} and yn=n{y}_{{n}}={n}. Clearly xn→∞{x}_{{n}}\to\infty and yn→∞{y}_{{n}}\to\infty. Then xnyn=n→∞\frac{{{x}_{{n}}}}{{{y}_{{n}}}}={n}\to\infty.

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

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

In this case we say that when xn→±∞{x}_{{n}}\to\pm\infty and yn→±∞{y}_{{n}}\to\pm\infty, expression xnyn\frac{{{x}_{{n}}}}{{{y}_{{n}}}} is indeterminate form of type ∞∞\frac{\infty}{\infty}.

Case 3. Now consider product xnyn{x}_{{n}}{y}_{{n}}. Here we will have indetermination when xn→0{x}_{{n}}\to{0} and yn→±∞{y}_{{n}}\to\pm\infty or vice versa.

If we take bn=1yn{b}_{{n}}=\frac{{1}}{{y}_{{n}}} then bn→0{b}_{{n}}\to{0}. So, xnyn=xn1yn=xnbn{x}_{{n}}{y}_{{n}}=\frac{{{x}_{{n}}}}{{\frac{{1}}{{y}_{{n}}}}}=\frac{{{x}_{{n}}}}{{{b}_{{n}}}} and we again have indeterminate form of type 00\frac{{0}}{{0}}.

Therefore, when xn→0{x}_{{n}}\to{0} and yn→±∞{y}_{{n}}\to\pm\infty we say that expression xnyn{x}_{{n}}{y}_{{n}} is indeterminate form of type 0⋅∞0\cdot\infty. This form can be transformed into indeterminate form of type 00\frac{{0}}{{0}}.

Case 4. Finally consider sum xn+yn{x}_{{n}}+{y}_{{n}}. Here we will have indeterminate form when xn{x}_{{n}} and yn{y}_{{n}} approach infinity with with diffferent signs.

Let xn=2n{x}_{{n}}={2}{n} and yn=−n{y}_{{n}}=-{n}. Clearly xn→∞{x}_{{n}}\to\infty and yn→−∞{y}_{{n}}\to-\infty. Then xn+yn=2n−n=n>∞{x}_{{n}}+{y}_{{n}}={2}{n}-{n}={n}>\infty.

Now let xn=n{x}_{{n}}={n} and yn=−2n{y}_{{n}}=-{2}{n}. Clearly xn→∞{x}_{{n}}\to\infty and yn−∞{y}_{{n}}-\infty. Then xn+yn=n−2n=−n→−∞{x}_{{n}}+{y}_{{n}}={n}-{2}{n}=-{n}\to-\infty.

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

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

In this case we say that when xn→+∞{x}_{{n}}\to+\infty and yn→−∞{y}_{{n}}\to-\infty, expression xn+yn{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 xn{x}_{{n}} and yn{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 xn=3n2−5n{x}_{{n}}={3}{{n}}^{{2}}-{5}{n}. Find limit of this sequence.

Since 3n2→∞{3}{{n}}^{{2}}\to\infty and −5n→−∞-{5}{n}\to-\infty we have indeterminate form of type ∞−∞\infty-\infty.

To handle it, let's perform algebraic manipulations: xn=n2(3−5n){x}_{{n}}={{n}}^{{2}}{\left({3}-\frac{{5}}{{n}}\right)}.

Now, since n2→∞{{n}}^{{2}}\to\infty and 3−5n→3{3}-\frac{{5}}{{n}}\to{3} then xn→∞{x}_{{n}}\to\infty.

Example 2. Let xn=−4n3+5n2{x}_{{n}}=-{4}{{n}}^{{3}}+{5}{{n}}^{{2}}. Find limit of this sequence.

Since −4n3→−∞-{4}{{n}}^{{3}}\to-\infty and 5n2→∞{5}{{n}}^{{2}}\to\infty we have indeterminate form of type ∞−∞\infty-\infty.

To handle it, let's perform algebraic manipulations: xn=n3(−4+5n){x}_{{n}}={{n}}^{{3}}{\left(-{4}+\frac{{5}}{{n}}\right)}.

Now, since n3→∞{{n}}^{{3}}\to\infty and −4+5n→−4-{4}+\frac{{5}}{{n}}\to-{4} then xn→−∞{x}_{{n}}\to-\infty.

Example 3. Let xn=a0nk+a1nk−1+…+ak−1n+ak{x}_{{n}}={a}_{{0}}{{n}}^{{k}}+{a}_{{1}}{{n}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{n}+{a}_{{k}}, where a0,a1,…ak{a}_{{0}},{a}_{{1}},\ldots{a}_{{k}} are constants. Find limit of this sequence.

This is generalization of above two examples. If all coefficients a0,a1,…,ak{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: xn=nk(a0+a1n+…+ak−1nk−1+aknk){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 nk→∞{{n}}^{{k}}\to\infty and a0+a1n+…+ak−1nk−1+aknk→a0{a}_{{0}}+\frac{{{a}_{{1}}}}{{n}}+\ldots+\frac{{{a}_{{{k}-{1}}}}}{{{{n}}^{{{k}-{1}}}}}+\frac{{{a}_{{k}}}}{{{{n}}^{{k}}}}\to{a}_{{0}} then xn→∞{x}_{{n}}\to\infty if a0>0{a}_{{0}}>{0} and xn→−∞{x}_{{n}}\to-\infty if a0<0{a}_{{0}}<{0}.

Example 4. Let xn=3n2−5n7n+3{x}_{{n}}=\frac{{{3}{{n}}^{{2}}-{5}{n}}}{{{7}{n}+{3}}}. Find limit of this sequence.

Since 3n2−5n→∞{3}{{n}}^{{2}}-{5}{n}\to\infty and 7n+3→∞{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{n} raised to the greatest degree in numerator and denominator (in this case n2{{n}}^{{2}}): xn=n2(3−5n)n2(7n+3n2)=3−5n7n+3n2{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 3−5n→3{3}-\frac{{5}}{{n}}\to{3} and 7n+3n2→0\frac{{7}}{{n}}+\frac{{3}}{{{n}}^{{2}}}\to{0} then xn→∞{x}_{{n}}\to\infty.

Example 5. Let xn=6n4−3n28n7+3{x}_{{n}}=\frac{{{6}{{n}}^{{4}}-{3}{{n}}^{{2}}}}{{{8}{{n}}^{{7}}+{3}}}. Find limit of this sequence.

Since 6n4−3n2→∞{6}{{n}}^{{4}}-{3}{{n}}^{{2}}\to\infty and 8n7+3n→∞{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{n} raised to the greatest degree in numerator and denominator (in this case n7{{n}}^{{7}}): xn=n7(6n3−3n5)n7(8+3n6)=6n3−3n58+3n6{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 6n3−3n5→0\frac{{6}}{{{n}}^{{3}}}-\frac{{3}}{{{n}}^{{5}}}\to{0} and 8+3n6→8{8}+\frac{{3}}{{{n}}^{{6}}}\to{8} then xn→0{x}_{{n}}\to{0}.

Example 6. Let xn=3n2−5n7n2+3{x}_{{n}}=\frac{{{3}{{n}}^{{2}}-{5}{n}}}{{{7}{{n}}^{{2}}+{3}}}. Find limit of this sequence.

Since 3n2−5n→∞{3}{{n}}^{{2}}-{5}{n}\to\infty and 7n2+3→∞{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{n} raised to the greatest degree in numerator and denominator (in this case n2{{n}}^{{2}}): xn=n2(3−5n)n2(7+3n2)=3−5n7+3n2{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 3−5n→3{3}-\frac{{5}}{{n}}\to{3} and 7+3n2→7{7}+\frac{{3}}{{{n}}^{{2}}}\to{7} then xn→37{x}_{{n}}\to\frac{{3}}{{7}}.

Example 7. Let xn=a0nk+a1nk−1+…+ak−1n+akb0nm+b1nm−1+..+bm−1n+bm{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 a0,a1,…,ak{a}_{{0}},{a}_{{1}},\ldots,{a}_{{k}} and b0,b1,…,bm{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 nk{{n}}^{{k}} from numerator and nm{{n}}^{{m}} from denominator: xn=nk(a0+a1n+…+aknk)nm(b0+b1n+…+bmnm)=nk−m(a0+a1n+…+aknkb0+b1n+…+bmnm){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 a0b0\frac{{a}_{{0}}}{{b}_{{0}}}. If k=m{k}={m} then nk−m=1→1{{n}}^{{{k}-{m}}}={1}\to{1} and xn→a0b0{x}_{{n}}\to\frac{{{a}_{{0}}}}{{{b}_{{0}}}}. If k>m{k}>{m} then nk−m→∞{{n}}^{{{k}-{m}}}\to\infty and xn→∞{x}_{{n}}\to\infty (or −∞-\infty, sign depends on sign of a0b0\frac{{{a}_{{0}}}}{{{b}_{{0}}}}). If k<m{k}<{m} then nk−m→0{{n}}^{{{k}-{m}}}\to{0} and xn→0{x}_{{n}}\to{0}.

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

We have indeterminate form of type ∞−∞\infty-\infty.

We have that 0<(n+1)k−nk=nk((1+1k)k−1)<nk((1+1n)−1)=nk−1=1n1−k{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<(n+1)k−nk<1n1−k{0}<{{\left({n}+{1}\right)}}^{{k}}-{{n}}^{{k}}<\frac{{1}}{{{{n}}^{{{1}-{k}}}}}.

Since 1n1−k→0\frac{{1}}{{{{n}}^{{{1}-{k}}}}}\to{0} then by Squeeze Theorem (n+1)k−nk→0{{\left({n}+{1}\right)}}^{{k}}-{{n}}^{{k}}\to{0}.

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

According to example 8 n+1−n→0\sqrt{{{n}+{1}}}-\sqrt{{{n}}}\to{0} that's why we have intedeterminate form of type ∞⋅0\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 n+1+n\sqrt{{{n}+{1}}}+\sqrt{{{n}}}:

xn=n(n+1−n)(n+1+n+1)n+1+n=n((n+1)2−(n)2)n+1+n=n(n+1−n)n+1+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}}}}}=

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

Now, factor out n\sqrt{{{n}}}: xn=nn(1+1n+1)=11+1n+1{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<1+1n<1+1n{1}<\sqrt{{{1}+\frac{{1}}{{n}}}}<{1}+\frac{{1}}{{n}} and 1+1n→1{1}+\frac{{1}}{{n}}\to{1} then by squeeze theorem 1+1n→1\sqrt{{{1}+\frac{{1}}{{n}}}}\to{1}.

That's why xn=11+1n−1→11+1=12{x}_{{n}}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{n}}}}-{1}}}\to\frac{{1}}{{{1}+{1}}}=\frac{{1}}{{2}}.

Example 10. Find limits of xn=nn2+n{x}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{n}}}}}, yn=nn2+1{y}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{1}}}}} and zn=1n2+1+1n2+2+…+1n2+n{z}_{{n}}=\frac{{1}}{{\sqrt{{{{n}}^{{2}}+{1}}}}}+\frac{{1}}{{\sqrt{{{{n}}^{{2}}+{2}}}}}+\ldots+\frac{{1}}{{\sqrt{{{{n}}^{{2}}+{n}}}}}.

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

Since xn=nn2+n=nn1+1n=11+1n{x}_{{n}}=\frac{{n}}{{\sqrt{{{{n}}^{{2}}+{n}}}}}=\frac{{n}}{{{n}\sqrt{{{1}+\frac{{1}}{{n}}}}}}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{n}}}}}} and 1+1n→1\sqrt{{{1}+\frac{{1}}{{n}}}}\to{1} then xn→1{x}_{{n}}\to{1}.

Also, since yn=nn2+1=nn1+1n2=11+1n2{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 1+1n2→1\sqrt{{{1}+\frac{{1}}{{{n}}^{{2}}}}}\to{1} then yn→1{y}_{{n}}\to{1}.

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

1n2+n+1n2+n+…+1n2+n<zn<1n2+1+1n2+1+…+1n2+1\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

nn2+n<zn<nn2+1\frac{{n}}{\sqrt{{{{n}}^{{2}}+{n}}}}<{z}_{{n}}<\frac{{n}}{\sqrt{{{{n}}^{{2}}+{1}}}} i.e. xn<zn<yn{x}_{{n}}<{z}_{{n}}<{y}_{{n}}.

Since xn→1{x}_{{n}}\to{1} and yn→1{y}_{{n}}\to{1} then by Squeeze Theorem zn→1{z}_{{n}}\to{1}.

Example 11. Suppose we are given m{m} positive numbers a1,a2,…,am{a}_{{1}},{a}_{{2}},\ldots,{a}_{{m}}. Find limit of xn=a1n+a2n+…+amnn{x}_{{n}}={\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}.

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

An+0+…+0n<a1n+a2n+…+amnn<A+A+…+An{\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<a1n+a2n+…+amnn<Amn{A}<{\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}<{A}{\sqrt[{{n}}]{{{m}}}}.

Since mn→1{\sqrt[{{n}}]{{{m}}}}\to{1} then by Squeeze Theorem xn=a1n+a2n+…+amnn→A{x}_{{n}}={\sqrt[{{n}}]{{{{a}_{{1}}^{{n}}}+{{a}_{{2}}^{{n}}}+\ldots+{{a}_{{m}}^{{n}}}}}}\to{A}.