Evaluating Limits of Sequences Examples 1

We will now look at some more examples of evaluating the limit of a sequence. More examples can be found on the Evaluating Limits of Sequences and Evaluating Limits of Sequences Examples 2 pages.

Example 1

Evaluate the limit $\lim_{n \to \infty} e^{1/n}$.

We first note that $\lim_{n \to \infty} \frac{1}{n} = 0$. Since the exponential function is continuous at $0$, it follows that:

(1)

\begin{align} \lim_{n \to \infty} e^{1/n} = e^{\lim_{n \to \infty} 1/n} = e^{0} = 1 \end{align}

Example 2

Evaluate the limit $\lim_{n \to \infty} \tan \left ( \frac{2n\pi}{1 + 8n} \right )$.

We note that $\lim_{n \to \infty} \frac{2n\pi}{1 + 8n} = \lim_{n \to \infty} \frac{2\pi}{\frac{1}{n} + 8} = \frac{2\pi}{8} = \frac{\pi}{4}$. Since the trigonometric tangent function is continuous at $\frac{pi}{4}$ it follows that:

(2)

\begin{align} \quad \lim_{n \to \infty} \tan \left ( \frac{2n\pi}{1 + 8n} \right ) = \tan \left ( \lim_{n \to \infty} \frac{2n\pi}{1 + 8n} \right ) = \tan \left ( \frac{\pi}{4} \right ) = 1 \end{align}

Example 3

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

Recall that from an earlier theorem that if $\lim_{n \to \infty} \mid a_n \mid = 0$ then $\lim_{n \to \infty} a_n = 0$. Let's first take the limit of the absolute value of our sequence above:

(3)

\begin{align} \lim_{n \to \infty} \biggr \rvert \frac{(-1)^n}{n} \biggr \rvert = \lim_{n \to \infty} \frac{1}{n} = 0 \end{align}

Therefore since $\lim_{n \to \infty} \biggr \rvert \frac{(-1)^n}{n} \biggr \rvert = 0$, it follows that $\lim_{n \to \infty} \frac{1}{n} = 0$.

Example 4

Evaluate the limit $\lim_{n \to \infty} \frac{(n!)^2}{(2n)!}$.

Expanding the factorial out we obtain that:

(4)

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

And therefore by the product law for the limit of a sequence:

(5)

\begin{align} \quad = \left ( \lim_{n \to \infty} \frac{1}{n + 1} \right )\left ( \lim_{n \to \infty} \frac{2}{n + 2} \right ) ... \left ( \lim_{n \to \infty} \frac{n -1}{2n - 1} \right )\left ( \lim_{n \to \infty} \frac{n}{2n} \right ) = 0 \cdot 0 \cdot ... \cdot 0 \cdot \frac{1}{2} = 0 \end{align}