When to simply plug in infinity when evaluating limits to infinity. Please don't destroy me as I am teaching myself calculus.
Anyway, my question pertains to this problem:
The problem with solution.
This is the solution to the limit of $x-$ $\sqrt{x^2-7}$ as $x$ approaches -$\infty$.
My question is, when is it okay to just plug in -$\infty$ like they did in the solution?
Also, when I did the problem, I got an answer of 0. Here is my step by step solution:
My solution
Feel free to critique it and give me and tips, tricks, and advice.
Thank you!
Edit: I made an error. The denominator in my solution should be a 0 as -1 +1 = 0.My final result based on the work I did should be 0/0.  Nevertheless, my answer is still incorrect.
Your basic building blocks for infinite limits are these:
$\lim_{x \to \infty} c = c$
$\lim_{x \to \infty} x = \infty$ and $\lim_{x \to -\infty} x = -\infty$
$\lim_{x \to \infty} 1/x = 0$ and so $\lim_{x \to \infty} 1/x^n = 0$ for $n &gt; 0$
The way I would have rearranged it is
$$x-\sqrt{x^2-7} = x - |x|\sqrt{1 - 7/x^2}$$ and noting that we are headed into the negative numbers, $|x| = -x$, we have
$$x+x\sqrt{1-7/x^2} = x\big(1+ \sqrt{1-7/x^2}\big) $$
and using the product rule for limits,
$$\lim_{x \to -\infty} x\big(1+ \sqrt{1-7/x^2}\big) = \lim_{x \to -\infty} x\cdot \lim_{x \to -\infty}\big(1+ \sqrt{1-7/x^2}\big)$$
I gathered all the finite parts so the last limit in the expression is $2$, while the first limit gives $-\infty$ so the final answer is $-\infty$.  So I don't plug in $\infty$ until the last possible moment, hoping that they will all cancel out beforehand without my having to guess what indeterminate forms such as $\infty - \infty$ might equal in this particular problem.
The limit of a sum is the sum of the limits (if it exists):
$$\lim_{x \to a} f(x) = b \quad\text{and}\quad \lim_{x \to a} g(x) = c \quad\text{implies}\quad \lim_{x \to a} f(x) + g(x) = b + c$$
Let $a = -\infty$, $f(x) = x$, $g(x) = -\sqrt{x^2-7}$. Then $b = c = -\infty$. Thus the answer is $-\infty$.
It is not always ok to simply plug in $\infty$. For example,
$$\lim_{x \to \infty} (x + a - x) = \lim_{x \to \infty} a = a$$
but
$$\lim_{x \to \infty} (x + a) - \lim_{x \to \infty} x = \infty - \infty$$
where $\infty - \infty$ is an indeterminate form. For a formalization of the idea of when you can &quot;plug in&quot; $\infty$, see the extended real number line:

By adjoining the elements $+\infty$ and $-\infty$ to $\mathbb{R}$, it enables a formulation of a &quot;limit at infinity&quot;, with topological properties similar to those for $\mathbb{R}$.
To make things completely formal, the Cauchy sequences definition of $\mathbb{R}$ allows defining $+\infty$ as the set of all sequences $\left\{ a_n \right\}$ of rational numbers, such that every $M \in \mathbb{R}$ is associated with a corresponding $N \in \mathbb{N}$ for which $a_n &gt; M$ for all $n &gt; N$. The definition of $-\infty$ can be constructed similarly.

Note the permitted arithmetic operations.