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.
 A: 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 $n > 0$
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.
A: A good answer here gives you the rules as of when should you "plug in infinity". I think an important way to look at it is to think when you must not "plug in infinity", and the reason why. Those cases are just examples, but you can use them to judge for yourself in similar situations.
Those are wrong:
$$
\lim_{x\to\infty} x - x = \infty - \infty \\
\lim_{x\to\infty} x - 4x =  \infty - \infty \\
\lim_{x\to\infty} e^{x} - x = \infty - \infty \\
$$
Note that the first limit equals zero, the other two approach negative and positive infinity respectively, so "plugging infinity in" does not preserve the true meaning of the expression. This is why we do not like expressions of the form $\infty - \infty$
Those are wrong as well:
$$
\lim_{x\to\infty} \frac{x}{x} = \frac{\infty}{\infty} \\
\lim_{x\to\infty} \frac{x}{e^{x}} = \frac{\infty}{\infty}\\
$$
And the reasoning is similar -first limit goes to 1, second goes to 0. So avoid stuff like $$\frac{\infty}{\infty}$$
For similar reasons, you do not want to plug in infinity and end up having an expression like this one:
$$0 \cdot \infty$$
So general do-not-do:
Do not "plug in infinity" if the expression you get is one of the following:
$$
\frac{\infty}{\infty} \\
\infty \cdot 0 \\
\infty - \infty
$$
A: 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 "plug in" $\infty$, see the extended real number line:

By adjoining the elements $+\infty$ and $-\infty$ to $\mathbb{R}$, it enables a formulation of a "limit at infinity", 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 $a_n > M$ for all $n > N$. The definition of $-\infty$ can be constructed similarly.

Note the permitted arithmetic operations.
