What is the "formal" way of proving Limit as x approaches negative infinity of f(x) Where f(x) = sqrt(5-x)
I know it's positive infinity but in order to get that I had to "operate" on infinity which is not allowed. Is there a different way to do this problem such that operating on infinity does not occur?
Show that for any $M>0,$ there is an $N_M<0$ such that for $x<N_M$ we have $f(x)>M.$
This is the (usual) definition of $f(x)$ increasing without bound as $x$ decreases without bound.
Note that "infinity" never came up. The symbol $\infty$ is in large part just notational shorthand for unboundedness.