How I resolve this limit?

${\lim}_{n \to + \infty} n x {\left(1 - x\right)}^{n}$ where $x \in \left[0 , 1\right]$

1 Answer
Jul 10, 2018

$0$

Explanation:

Let's begin with the boundary cases:

If $x = 0$, the expression becomes

${\lim}_{n \setminus \to \setminus \infty} n \cdot 0 {\left(1 - 0\right)}^{n} = {\lim}_{n \setminus \to \setminus \infty} 0 = 0$

If $x = 1$, the expression becomes

${\lim}_{n \setminus \to \setminus \infty} n \cdot 1 {\left(1 - 1\right)}^{n} = {\lim}_{n \setminus \to \setminus \infty} 0 = 0$

For every $x \setminus \in \left(0 , 1\right)$, we can take the $x$ out of the limit, since it doesn't depend on $n$:

${\lim}_{n \setminus \to \setminus \infty} n x {\left(1 - x\right)}^{n} = x {\lim}_{n \setminus \to \setminus \infty} n {\left(1 - x\right)}^{n}$

Since $x \setminus \in \left(0 , 1\right)$, $1 - x$ also belongs to $\left(0 , 1\right)$. So, we have a limit like

$x {\lim}_{n \setminus \to \setminus \infty} n \cdot {y}^{n} = x \cdot \left(\infty \cdot 0\right)$

A limit like $\setminus \infty \cdot 0$ can't be answered right away, but we see that the part going to infinity is linear ($n$), whereas the part going to zero is exponential (${y}^{n}$, for $0 < y < 1$). So, this part dominates, and the whole expression goes to zero.