Find the limit of function $f_{n}(x) = nx^{n}$ Show that $\lim_{n \rightarrow \infty} nx^{n} = 0$ for $x \in [0, 1), n \in \mathbb{N}$

$\lim_{n \rightarrow \infty} nx^{n}$

$\Rightarrow \lim_{n \rightarrow \infty} n \cdot \lim_{n \rightarrow \infty} x^{n}$

$\Rightarrow \infty * 0$

$\Rightarrow 0$

Is this correct?

Now show that $\lim_{n \rightarrow \infty} \int_{0}^{1} f_{n}(x) dx = 1$

$\lim_{n \rightarrow \infty} \int_{0}^{1} nx^{n} dx$

$\Rightarrow \lim_{n \rightarrow \infty} \frac{n}{n + 1} x^{n + 1}$

(then plug in $1$ and $0$ into $x$)

$\Rightarrow \lim_{n \rightarrow \infty} \frac{1}{2}x^2$

Which is not $=1$, where did I go wrong?
As others have noted, even if $0\cdot\infty$ was defined (and it isn't), you still couldn't just break up the product inside the limit. Instead, you could apply L'Hoptial's rule to get:
$$\lim_{n\to\infty}\frac{n}{x^{-n}} = \lim_{n\to\infty}\frac{1}{-x^{-n}\log x}=0$$
$$
\lim_{n\rightarrow \infty}\left(nx^n \right)=
\lim_{n\rightarrow \infty}\left(n \over x^{-n} \right)=
\lim_{n\rightarrow \infty}\left(1 \over (-n)x^{-n-1} \right)=
\lim_{n\rightarrow \infty}\left(-x^{n+1} \over n \right)=0\;  (for\;\; 0\leq x&lt;1)
$$