Theorems for and Examples of Computing Limits of Sequences

Evaluating limits of sequences using limits of functions

The limit of a functiion is the limit of continuous values of the function, while the limit of a sequence is a limit of discrete numbers.  Nonetheless, we recognize that if a continuous function passes through all numbers of a sequence, then the convergence or divergence of the sequence matches the convergence or divergence of the function.  To see this, just picture the graph of a sequence (earlier in this module) and draw the curve through the dots, letting the curve be $f(x)$, and hence $f(n)=a_n$ for the positive integers $n$.

Example 1:  By the theorem, since $\displaystyle\lim_{x\to\infty}\frac{1}{x^r}=0$ when $r>0$, $\displaystyle\lim_{n\to\infty}\frac{1}{n^r}=0$ when $r>0$.  Learn this example.
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Example 2:  Evaluate $\displaystyle\lim_{n\to\infty}\frac{2n}{3n-4}$.

Solution 2:  By dividing the top and the bottom by $n$, we get $\displaystyle\lim_{n\to\infty}\frac{2}{3-\tfrac 4 n}$.   By the previous example, $\tfrac 4 n$ converges to 0, so we get $\displaystyle\lim_{n\to\infty}\frac{2n}{3n-4}=\frac{2}{3}$.
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Example 3:  Evaluate $\displaystyle\lim_{n\to\infty}\frac{5n}{e^n}$.  DO:  To solve this limit, first compute $\displaystyle\lim_{x\to\infty}\frac{5x}{e^x}$ before reading on.



Solution 3:  We use l'Hospital's Rule. $\displaystyle\lim_{x\to\infty}\frac{5x}{e^x}\underset{\fbox{$\tfrac \infty \infty\,,\, \text{l'H}$}}{=}\lim_{x\to\infty}\frac{5}{e^x}=0$.  Thus $\displaystyle\lim_{n\to\infty}\frac{5n}{e^n}=0$ by Theorem 1.
Notice that we cannot take a derivative of a sequence, since a derivative exists only if the function is continuous, and sequences are not continuous.  Yet, because of our theorem, we can use l'Hopital's rule on the associated (continuous) $f(x)$ and compute our limit by taking derivatives.  We get the limit of the sequence by evaluating the limit of the function.

Alternating sign sequences

Example 4:  By Theorem 1, since $\displaystyle\lim_{x\to\infty}r^x=0$ when $0<r<1$ (remember exponential functions?), $\displaystyle\lim_{n\to\infty}r^n=0$ when $0<r<1$.  By Theorem 2, this extends to $-1<r<1$.  If $r>1$ or $r\le-1$, the limit diverges.  (DO:  what happens when $r=1$?)  Summarizing:
$\displaystyle\lim_{n\to\infty}r^n \left\{\begin{array}{ll}\text{converges }&\text{ if }-1<r\le 1\\\text{diverges}&\text{otherwise}\end{array}\right.$  Learn this.  We will use it frequently.

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Example 5:  Let $ a_n=(-1)^n\frac{3^{n+2}}{5^n}$.  We will take the the absolute value of our sequence, which here simply means ignoring the alternating sign $(-1)^n$, and take a limit to see if we get 0.  Rewrite to get $|a_n|=3^2\left(\frac{3}{5}\right)^n=9\left(\frac{3}{5}\right)^n$.  We evaluate
$\displaystyle\lim_{n\to\infty}|a_n|= \lim_{n\to\infty}\frac{3^{n+2}}{5^n}=9\left(\frac{3}{5}\right)^n=9\cdot 0=0$, by Example 4 with $r=\tfrac 3 5$, since $-1<r<1$.  Therefore,
$\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}(-1)^n\frac{3^{n+2}}{5^n}=0$ by Theorem 2.