Use a table of values of f(x) to estimate the value of the limit to four decimal places.
idk if I am just not knowing how to do this, but if someone could walk me through this I would be more than greatful!
$\displaystyle \lim_{x\rightarrow-\infty} \sqrt{3x^2+4x+5}-\sqrt{3x^2+3x+7}$
$\displaystyle \lim_{x\rightarrow-\infty} \sqrt{3x^2+4x+5}-\sqrt{3x^2+3x+7}\frac{(\sqrt{3x^2+4x+5}+\sqrt{3x^2 +3x+7})}{(\sqrt{3x^2+4x+5}-\sqrt{3x^2+3x+7})}$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{3x^2+4x+5-(3x^2+3x+7)}{\sqrt{3x^2+4x+5}+\sqrt{3x^2+3x+7}}$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{x-2}{\sqrt{3x^2+4x+5}+\sqrt{3x^2+3x+7}}$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{-(x-2)/|x|}{(\sqrt{3x^2+4x+5}+\sqrt{3x^2+3x+7})/|x|}$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{-(1-\frac{2}{x})}{\sqrt{3+\frac{4}{x}+\frac{5}{x^2}}+\ sqrt{3+\frac{3}{x}+\frac{7}{x^2}}}$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{-(1-0)}{\sqrt{3+0+0}+\sqrt{3+0+0}}$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{3}+\sqrt{3}}=\frac{-1}{2\sqrt{3}}$
Holy mother I can't believe I typed that all out.
There was a solution already posted similar to this question here: