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:
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 $n > 0$
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.