I'm to find the limit (if it exists), if it does not then explain.
$$ \lim_{x\to4^-} \dfrac{\sqrt{x}-2}{x-4}.$$
So it's my understanding that if $x$ approaches 4 from the left, I can use a value like $3.999$? Is there a more algebraic way to do this? Maybe I should use conjugates?
On
You could certainly use $3.999$, or $3.99999$, or something with even more "$9$'s". If you want to go that route try $3.999$, and then $3.9999$ and see if the answers are reasonably close. Keep adding "$9$'s" until you can reasonably predict the limit.
L'Hopital's rule will also work. (If you are unfamiliar with this rule ignore this method.) To use L'Hopital take the derivative of the top and the derivative of the bottom, then take the limit of that new fraction. Here your question becomes: $$\lim_{x\to4^-}\dfrac{1}{2\sqrt{x}}$$ and you can caculate the limit from there.
Conjugates will also work. Multiply the entire limit by $\frac{\sqrt{x}+2}{\sqrt{x}+2}$ to get:$$\lim_{x\to4^-}\dfrac{x-4}{(x-4)(\sqrt{x}+2)} = \lim_{x\to4^-}\dfrac{1}{\sqrt{x}+2}$$ and again, proceed from there.
Hint: $x-4=(\sqrt{x}-2)(\sqrt{x}+2)$, then simplify...