Evaluating Limits of Sequences

We will now look at some examples of evaluating limits of sequences using various approaches and theorems that we have already looked at. More examples can be found on the Evaluating Limits of Sequences Examples 1 and Evaluating Limits of Sequences Examples 2 pages.

Example 1

Evaluate the limit $\lim_{n \to \infty} \frac{n^2 - 4}{n + 5}$.

This limit should be relatively easy to evaluate since all we have to do is divide each part of the fraction by $n$ to get that:

(1)

\begin{align} \quad \lim_{n \to \infty} \frac{n^2 - 4}{n + 5} = \lim_{n \to \infty} \frac{n - \frac{4}{n}}{1 + \frac{5}{n}} = \lim_{n \to \infty} n = \infty \end{align}

Example 2

Evaluate the limit $\lim_{n \to \infty} \frac{e^n - e^{-n}}{e^n + e^{-n}}$.

Once again, we will divide each part of the fraction, but this time by $e^n$ and noting that $\lim_{n \to \infty} e^{-2n} = 0$, we get that:

(2)

\begin{align} \quad \lim_{n \to \infty} \frac{e^n - e^{-n}}{e^n + e^{-n}} = \lim_{n \to \infty} \frac{1 - \frac{e^{-n}}{e^n}}{1 + \frac{e^{-n}}{e^{n}}} = \lim_{n \to \infty} \frac{1 - e^{-2n}}{1 + e^{-2n}} = 1 \end{align}

Example 3

Evaluate the limit $\lim_{n \to \infty} \sqrt{n^2 + n} - \sqrt{n^2 - 1}$.

This time we will multiply the equation of our sequence by the conjugate of our sequence and divide by the conjugate to obtain:

(3)

\begin{align} \quad \lim_{n \to \infty} \left ( \sqrt{n^2 + n} - \sqrt{n^2 - 1} \right )= \lim_{n \to \infty} \left ( \sqrt{n^2 + n} - \sqrt{n^2 - 1} \cdot \frac{\sqrt{n^2 + n} - \sqrt{n^2 - 1}}{\sqrt{n^2 + n} - \sqrt{n^2 - 1}} \right ) = \lim_{n \to \infty} \frac{n + 1}{\sqrt{n^2 + n} + \sqrt{n^2 - 1}} \end{align}

Now we will factor an $n$ out of the radicals of the denominator, divide each term by $n$, and note that $\lim_{n \to \infty} \frac{1}{n} = 0$, we get that:

(4)

\begin{align} = \lim_{n \to \infty} \frac{n + 1}{n\sqrt{1 + \frac{1}{n}} + n\sqrt{1 - \frac{1}{n^2}}} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{\sqrt{1 + \frac{1}{n}} + \sqrt{1 - \frac{1}{n}}} = \frac{1}{\sqrt{1}+\sqrt{1}} = \frac{1}{2} \end{align}

Example 4

Evaluate the limit $\lim_{n \to \infty} (-1)^n \frac{n}{n^3 + 1}$.

Dividing each part of the fraction by $n^3$ we get that:

(5)

\begin{align} \lim_{n \to \infty} (-1)^n \frac{n}{n^3 + 1} = \lim_{n \to \infty} (-1)^n \frac{\frac{1}{n^2}}{1 + \frac{1}{n^3}} = 0 \end{align}