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Algebraic limit laws

Given sequences { a n } and { b n } and any real number c , if there exist constants A and B such that lim n a n = A and lim n b n = B , then

  1. lim n c = c
  2. lim n c a n = c lim n a n = c A
  3. lim n ( a n ± b n ) = lim n a n ± lim n b n = A ± B
  4. lim n ( a n · b n ) = ( lim n a n ) · ( lim n b n ) = A · B
  5. lim n ( a n b n ) = lim n a n lim n b n = A B , provided B 0 and each b n 0 .

Proof

We prove part iii.

Let ϵ > 0 . Since lim n a n = A , there exists a constant positive integer N 1 such that for all n N 1 . Since lim n b n = B , there exists a constant N 2 such that | b n B | < ε / 2 for all n N 2 . Let N be the largest of N 1 and N 2 . Therefore, for all n N ,

| ( a n + b n ) ( A + B ) | | a n A | + | b n B | < ε 2 + ε 2 = ε .

The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence { 1 n 2 } . As shown earlier, lim n 1 / n = 0 . Similarly, for any positive integer k , we can conclude that

lim n 1 n k = 0 .

In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.

Determining convergence and finding limits

For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.

  1. { 5 3 n 2 }
  2. { 3 n 4 7 n 2 + 5 6 4 n 4 }
  3. { 2 n n 2 }
  4. { ( 1 + 4 n ) n }
  1. We know that 1 / n 0 . Using this fact, we conclude that
    lim n 1 n 2 = lim n ( 1 n ) . lim n ( 1 n ) = 0 .

    Therefore,
    lim n ( 5 3 n 2 ) = lim n 5 3 lim n 1 n 2 = 5 3.0 = 5 .

    The sequence converges and its limit is 5 .
  2. By factoring n 4 out of the numerator and denominator and using the limit laws above, we have
    lim n 3 n 4 7 n 2 + 5 6 4 n 4 = lim n 3 7 n 2 + 5 n 4 6 n 4 4 = lim n ( 3 7 n 2 + 5 n 4 ) lim n ( 6 n 4 4 ) = ( lim n ( 3 ) lim n 7 n 2 + lim n 5 n 4 ) ( lim n 6 n 4 lim n ( 4 ) ) = ( lim n ( 3 ) 7 · lim n 1 n 2 + 5 · lim n 1 n 4 ) ( 6 · lim n 1 n 4 lim n ( 4 ) ) = 3 7 · 0 + 5 · 0 6 · 0 4 = 3 4 .

    The sequence converges and its limit is −3 / 4 .
  3. Consider the related function f ( x ) = 2 x / x 2 defined on all real numbers x > 0 . Since 2 x and x 2 as x , apply L’Hôpital’s rule and write
    lim x 2 x x 2 = lim x 2 x ln 2 2 x Take the derivatives of the numerator and denominator. = lim x 2 x ( ln 2 ) 2 2 Take the derivatives again. = .

    We conclude that the sequence diverges.
  4. Consider the function f ( x ) = ( 1 + 4 x ) x defined on all real numbers x > 0 . This function has the indeterminate form 1 as x . Let
    y = lim x ( 1 + 4 x ) x .

    Now taking the natural logarithm of both sides of the equation, we obtain
    ln ( y ) = ln [ lim x ( 1 + 4 x ) x ] .

    Since the function f ( x ) = ln x is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,
    ln ( y ) = lim x [ ln ( 1 + 4 x ) x ] .

    Using properties of logarithms, we write
    lim x [ ln ( 1 + 4 x ) x ] = lim x x ln ( 1 + 4 x ) .

    Since the right-hand side of this equation has the indeterminate form · 0 , rewrite it as a fraction to apply L’Hôpital’s rule. Write
    lim x x ln ( 1 + 4 x ) = lim x ln ( 1 + 4 / x ) 1 / x .

    Since the right-hand side is now in the indeterminate form 0 / 0 , we are able to apply L’Hôpital’s rule. We conclude that
    lim x ln ( 1 + 4 / x ) 1 / x = lim x 4 1 + 4 / x = 4 .

    Therefore, ln ( y ) = 4 and y = e 4 . Therefore, since lim x ( 1 + 4 x ) x = e 4 , we can conclude that the sequence { ( 1 + 4 n ) n } converges to e 4 .
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Consider the sequence { ( 5 n 2 + 1 ) / e n } . Determine whether or not the sequence converges. If it converges, find its limit.

The sequence converges, and its limit is 0 .

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Recall that if f is a continuous function at a value L , then f ( x ) f ( L ) as x L . This idea applies to sequences as well. Suppose a sequence a n L , and a function f is continuous at L . Then f ( a n ) f ( L ) . This property often enables us to find limits for complicated sequences. For example, consider the sequence 5 3 n 2 . From [link] a. we know the sequence 5 3 n 2 5 . Since x is a continuous function at x = 5 ,

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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