# What is the limit of x^n?

Jul 13, 2015

${\lim}_{n \to \infty} {x}^{n}$ behaves in seven different ways according to the value of $x$

#### Explanation:

If $x \in \left(- \infty , - 1\right)$ then as $n \to \infty$, $\left\mid {x}^{n} \right\mid \to \infty$ monotonically, but alternates between positive and negative values. ${x}^{n}$ does not have a limit as $n \to \infty$.

If $x = - 1$ then as $n \to \infty$, ${x}^{n}$ alternates between $\pm 1$. So again, ${x}^{n}$ does not have a limit as $n \to \infty$.

If $x \in \left(- 1 , 0\right)$ then ${\lim}_{n \to \infty} {x}^{n} = 0$. The value of ${x}^{n}$ alternates between positive and negative values but $\left\mid {x}^{n} \right\mid \to 0$ is monotonically decreasing.

If $x = 0$ then ${\lim}_{n \to \infty} {x}^{n} = 0$. The value of ${x}^{n}$ is constant $0$ (at least for $n > 0$).

If $x \in \left(0 , 1\right)$ then ${\lim}_{n \to \infty} {x}^{n} = 0$ The value of ${x}^{n}$ is positive and ${x}^{n} \to 0$ monotonically as $n \to \infty$.

If $x = 1$ then ${\lim}_{n \to \infty} {x}^{n} = 1$. The value of ${x}^{n}$ is constant $1$.

If $x \in \left(1 , \infty\right)$ then as $n \to \infty$, then ${x}^{n}$ is positive and ${x}^{n} \to \infty$ monotonically. ${x}^{n}$ does not have a limit as $n \to \infty$.