Question

$\underset{x\to -\infty }{\lim }\left(\frac{6x^2-\cos 3x}{x^2+5}-\frac{5x^2+3}{\sqrt{x^6+2}}\right)=$

• A. 11
• B. 0
• C. -1
• D. 1

Answer 1

Answer: (A)

$\underset{x\to -\infty }{\lim }\left(\frac{6x^2-\cos 3x}{x^2+5}-\frac{5x^3+3}{\sqrt{x^6+2}}\right)$
$=\underset{x\to -\infty }{\lim }\left(\frac{6-\frac{\cos 3x}{x^2}}{1+\frac{5}{x^2}}-\frac{5+\frac{3}{x^3}}{\left(\frac{|x{|}^3}{x^3}\right)\sqrt{1+\frac{2}{x^6}}}\right)$
$\left[\because \underset{x\to -\infty }{\lim }\frac{\cos 3x}{x^2}=0\text{and}\underset{x\to -\infty }{\lim }\frac{|x{|}^3}{x^3}=-1\right]$
$=6+5=11$