Question

$\underset{x\to \infty }{lim}\frac{|3x^2+1|}{2x^2+1}=$

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $-\frac{3}{2}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is A $\frac{3}{2}$

The value of the expression, $3x^2+1$ is always $>0$ since $x^2>0$ for all $x$.
So, the expression simplifies to
${lim}_{x\to \infty }\frac{3x^2+1}{2x^2+1}$
Now dividing by $x^2$ we get,
${lim}_{x\to \infty }\frac{3+\frac{1}{x^2}}{2+\frac{1}{x^2}}$
Applying the limit we get,
$=\frac{3+0}{2+0}$ $=\frac{3}{2}$
Hence, option 'A' is correct.