Sequences and Their Properties: Learn It 4

Advanced Techniques for Sequence Limits

In the previous section, we learned the basic tools for finding sequence limits. Now we’ll explore more sophisticated techniques that handle challenging cases where basic limit laws aren’t enough.

L’Hôpital’s Rule for Sequences

Sometimes you’ll encounter sequences where both the numerator and denominator grow without bound, making it unclear what happens to their ratio. These indeterminate forms require special techniques.

We often need to analyze sequences featuring ratios where both parts increase without bound and it’s not immediately clear what the limit will be. Fortunately, we can use L’Hôpital’s Rule from our study of functions.

L’Hôpital’s Rule

Suppose [latex]f[/latex] and [latex]g[/latex] are differentiable functions over an open interval [latex](a, \infty)[/latex]. If either:

  • [latex]\underset{x\to \infty}{\lim}f(x) = 0[/latex] and [latex]\underset{x\to \infty}{\lim}g(x) = 0[/latex]
                                                                                          or
  • [latex]\underset{x\to \infty}{\lim}f(x) = \infty[/latex] (or [latex]-\infty[/latex]) and [latex]\underset{x\to \infty}{\lim}g(x) = \infty[/latex] (or [latex]-\infty[/latex])

Then: [latex]\underset{x\to \infty}{\lim}\frac{f(x)}{g(x)} = \underset{x\to \infty}{\lim}\frac{f'(x)}{g'(x)}[/latex]

(assuming the limit on the right exists or is [latex]\infty[/latex] or [latex]-\infty[/latex])

Consider the sequence [latex]\{\frac{(5{n}^{2}+1)}{{e}^{n}}\}[/latex]. Determine whether or not the sequence converges. If it converges, find its limit.

Watch the following video to see the worked solution to the above example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.1.2” here (opens in new window).

Limits with Continuous Functions

Recall that if [latex]f[/latex] is a continuous function at a value [latex]L[/latex], then [latex]f(x)\to f(L)[/latex] as [latex]x\to L[/latex]. This idea applies to sequences as well.

Consider the sequence [latex]{\sqrt{5-\frac{3}{n^2}}}[/latex]. We know that [latex]5-\frac{3}{n^2} \to 5[/latex]. Since [latex]\sqrt{x}[/latex] is continuous at [latex]x = 5[/latex]:

[latex]\underset{n\to \infty}{\lim}\sqrt{5-\frac{3}{n^2}} = \sqrt{\underset{n\to \infty}{\lim}(5-\frac{3}{n^2})} = \sqrt{5}[/latex]

continuous functions and convergent sequences

If sequence [latex]{a_n}[/latex] converges to [latex]L[/latex] and function [latex]f[/latex] is continuous at [latex]L[/latex], then:

[latex]\underset{n\to \infty}f(a_n) = f(\underset{n\to \infty}a_n) = f(L)[/latex]

Proof

Let [latex]>0[/latex]. Since [latex]f[/latex] is continuous at [latex]L[/latex], there exists [latex]\delta >0[/latex] such that [latex]|f(x)-f(L)|<\epsilon[/latex] if [latex]|x-L|<\delta[/latex]. Since the sequence [latex]\{{a}_{n}\}[/latex] converges to [latex]L[/latex], there exists [latex]N[/latex] such that [latex]|{a}_{n}-L|<\delta[/latex] for all [latex]n\ge N[/latex]. Therefore, for all [latex]n\ge N[/latex], [latex]|{a}_{n}-L|<\delta[/latex], which implies [latex]|f({a}_{n})\text{-}f(L)|<\epsilon[/latex]. We conclude that the sequence [latex]\{f({a}_{n})\}[/latex] converges to [latex]f(L)[/latex].

[latex]_\blacksquare[/latex]

A graph in quadrant 1 with points (a_1, f(a_1)), (a_3, f(a_3)), (L, f(L)), (a_4, f(a_4)), and (a_2, f(a_2)) connected by smooth curves.
Figure 4. Because [latex]f[/latex] is a continuous function as the inputs [latex]{a}_{1},{a}_{2},{a}_{3}\text{,}\ldots[/latex] approach [latex]L[/latex], the outputs [latex]f\left({a}_{1}\right),f\left({a}_{2}\right),f\left({a}_{3}\right)\text{,}\ldots[/latex] approach [latex]f\left(L\right)[/latex].

Determine whether the sequence [latex]\{\cos(\frac{3}{{n}^{2}})\}[/latex] converges. If it converges, find its limit.

The Squeeze Theorem for Sequences

Another powerful technique extends the Squeeze Theorem you learned for function limits. This method is particularly useful when you can’t find a sequence’s limit directly, but you can “sandwich” it between two sequences whose limits you do know.

squeeze theorem for sequences

Consider sequences [latex]\{{a}_{n}\}[/latex], [latex]\{{b}_{n}\}[/latex], and [latex]\{{c}_{n}\}[/latex]. Suppose there exists an integer [latex]N[/latex] such that

[latex]{a}_{n}\le {b}_{n}\le {c}_{n}\text{for all}n\ge N[/latex].

 

If there exists a real number [latex]L[/latex] such that

[latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=L=\underset{n\to \infty }{\text{lim}}{c}_{n}[/latex],

 

then [latex]\{{b}_{n}\}[/latex] converges and [latex]\underset{n\to \infty }{\text{lim}}{b}_{n}=L[/latex] (Figure 5).

 

A graph in quadrant 1 with the line y = L and the x-axis labeled as the n axis. Points are plotted above and below the line, converging to L as n goes to infinity. Points a_n, b_n, and c_n are plotted at the same n-value. A_n and b_n are above y = L, and c_n is below it.
Figure 5. Each term [latex]{b}_{n}[/latex] satisfies [latex]{a}_{n}\le {b}_{n}\le {c}_{n}[/latex] and the sequences [latex]\left\{{a}_{n}\right\}[/latex] and [latex]\left\{{c}_{n}\right\}[/latex] converge to the same limit, so the sequence [latex]\left\{{b}_{n}\right\}[/latex] must converge to the same limit as well.

Proof

Let [latex]\epsilon >0[/latex]. Since the sequence [latex]\{{a}_{n}\}[/latex] converges to [latex]L[/latex], there exists an integer [latex]{N}_{1}[/latex] such that [latex]|{a}_{n}-L|<\epsilon[/latex] for all [latex]n\ge {N}_{1}[/latex]. Similarly, since [latex]\{{c}_{n}\}[/latex] converges to [latex]L[/latex], there exists an integer [latex]{N}_{2}[/latex] such that [latex]|{c}_{n}-L|<\epsilon[/latex] for all [latex]n\ge {N}_{2}[/latex]. By assumption, there exists an integer [latex]N[/latex] such that [latex]{a}_{n}\le {b}_{n}\le {c}_{n}[/latex] for all [latex]n\ge N[/latex]. Let [latex]M[/latex] be the largest of [latex]{N}_{1},{N}_{2}[/latex], and [latex]N[/latex]. We must show that [latex]|{b}_{n}-L|<\epsilon[/latex] for all [latex]n\ge M[/latex]. For all [latex]n\ge M[/latex],

[latex]\text{-}\epsilon <\text{-}|{a}_{n}-L|\le {a}_{n}-L\le {b}_{n}-L\le {c}_{n}-L\le |{c}_{n}-L|<\epsilon[/latex].

 

Therefore, [latex]\text{-}\epsilon <{b}_{n}-L<\epsilon[/latex], and we conclude that [latex]|{b}_{n}-L|<\epsilon[/latex] for all [latex]n\ge M[/latex], and we conclude that the sequence [latex]\{{b}_{n}\}[/latex] converges to [latex]L[/latex].

[latex]_\blacksquare[/latex]

When to Use the Squeeze Theorem

Look for the Squeeze Theorem when:

  • The sequence involves trigonometric functions (since [latex]-1 \leq \sin x, \cos x \leq 1[/latex])
  • You have a sequence that oscillates but is bounded
  • Direct limit calculation seems difficult, but you can find upper and lower bounds

The key is identifying good “squeezing” sequences that are easier to analyze.

Use the Squeeze Theorem to find the limit of each of the following sequences.

  1. [latex]\{\frac{\cos{n}}{{n}^{2}}\}[/latex]
  2. [latex]\{{(-\frac{1}{2})}^{n}\}[/latex]