Find the limit.
$\lim _ { x \rightarrow 2 } \sqrt { \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } }$
A) 0
B) $- \frac { 4 } { 3 }$
C) $\frac { \sqrt { 17 } } { 2 }$
D) $\frac { 4 } { 3 }$
E) does not exist

Question

Quizplus · Accepted Answer

We can simplify the expression inside the square root using long division or synthetic division:

$equire{enclose}
\begin{array}{c|ccccc}
\multicolumn{2}{r}{2} &  &  &  &  \
\cline{2-6}
3x-2 & 4x^2 & 0x & 1 &  &  \
\multicolumn{2}{r}{4x^2} & -\frac{8}{3}x & & & \
\cline{2-3}
\multicolumn{2}{r}{0} & \frac{4}{3}x & 1 & & \
\multicolumn{2}{r}{} & \frac{4}{3}x & -\frac{8}{9} & & \
\cline{3-4}
\multicolumn{2}{r}{} &  & \frac{17}{9} & &
\end{array}$

Thus, we can write the original expression as

$\lim_{xightarrow 2} \sqrt{\frac{4x^2+1}{3x-2}} = \lim_{xightarrow 2}\sqrt{\frac{(3x-2)(4x^2+1)}{(3x-2)^2}} = \lim_{xightarrow 2}\sqrt{\frac{12x^2-8x+1}{9x^2-12x+4}}.$

At this point, we can apply L'Hopital's Rule, but it's simpler to notice that as $xightarrow 2,$ both the numerator and denominator of the fraction under the square root approach positive values. Thus, we can take the square root of the limits of the numerator and denominator separately. We get

$$\lim_{xightarrow 2}\sqrt{\frac{12x^2-8x+1}{9x^2-12x+4}} = \frac{\sqrt{\displaystyle\lim_{xightarrow 2}(12x^2-8x+1)}}{\sqrt{\displaystyle\lim_{xightarrow 2}(9x^2-12x+4)}} = \frac{\sqrt{45}}{\sqrt{20}} = \frac{\sqrt{45}}{2\sqrt{5}} = \frac{\sqrt{9}\sqrt{5}}{2\sqrt{5}} = \frac{3}{2}.$$

Therefore, the answer is $\boxed{	extbf{(C) }\frac{\sqrt{17}}{2}}.$

Find the Limit lim⁡x→24x2+13x−2\lim _ { x \rightarrow 2 } \sqrt { \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } }limx→2​3x−24x2+1​​

Find the Limit $\lim _ { x \rightarrow 2 } \sqrt { \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } }$