How do you solve this limit? I get sqrt(3)/4 as my final answer and apparently it is wrong. My work (and the question) is attached (although my handwriting is terrible).
Any help would be greatly appreciated!
Thanks in advance!
$\displaystyle \lim_{x\to-\infty}\frac{\sqrt{11+3x^2}}{2+4x}=\lim_{x\to-\infty}\frac{|x|\sqrt{\frac{11}{x^2}+3}}{x\left(\f rac{2}{x}+4\right)}=$
$\displaystyle =\lim_{x\to-\infty}\frac{-x\sqrt{\frac{11}{x^2}+3}}{x\left(\frac{2}{x}+4\rig ht)}=-\lim_{x\to-\infty}\frac{\sqrt{\frac{11}{x^2}+3}}{\frac{2}{x}+ 4}=-\frac{\sqrt{3}}{4}$