Let's say we have this: $$ \infty \cdot 0 + \infty$$ since $\infty\cdot0$ is not defined can we do this: $$ \infty \cdot(0 + 1) = \infty \cdot 1 = \infty $$ and therefore can I say $\infty\cdot0 + \infty = \infty$ ?
I guess there will be problem with doing arithmetic with $\infty\cdot0$ but I found this kind of interesting.
On
No you can't. It is still indeterminate form. The results may be different. Look at the examples:
Let $(a_n), (b_n), (c_n)$ be infinity sequence such that $a_n= n^4, b_n = \frac{-1}{n}, c_n = n$.
$$\lim_{n\to\infty}a_n\cdot b_n + c_n = \lim_{n\to\infty}n^4 \cdot\frac{-1}{n}+n = \lim_{n\to\infty} n(1-n^2) = -\infty$$
Or let $(a_n), (b_n), (c_n)$ be infinity sequence such that $a_n= n^2, b_n = \frac{-1}{n}, c_n = n$.
$$\lim_{n\to\infty}a_n\cdot b_n + c_n = \lim_{n\to\infty}n^2 \cdot\frac{-1}{n} + n = \lim_{n\to\infty} (-n) + n = 0$$
Of course in both above example $\lim_{n\to\infty} a_n = +\infty \wedge \lim_{n\to\infty} b_n = 0 \wedge \lim_{n\to\infty} c_n = +\infty$.
On
$$\infty \cdot (0+1) = \infty $$
is indeed correct, assuming by $\infty$ you mean the element of the extended real numbers.
The problem is that you can't say
$$ \infty \cdot (0+1) = \infty \cdot 0 + \infty \cdot 1 $$
The distributive law for extended real numbers is only valid where it is defined.
No, you can't do that. Such manipulations with $\infty$ aren't mathematically permitted. Perhaps more importantly for usage of $\infty$, they don't cooperate with limits, either. For example, let $f(x)=\frac{1}{x^2}$, $g(x)=x$, and $h(x)=\frac{1}{\sqrt{-x}}$. Then
$$\lim_{x \to 0^-} f(x) g(x) + h(x) = \lim_{x \to 0^-} \frac{1}{x} + \frac{1}{\sqrt{-x}} = -\infty$$
yet the limit is of the form $\infty \cdot 0 + \infty$.