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

To find the limit as $x$ approaches $2$, we need to evaluate the expression at $x=2$. However, direct substitution gives us $\frac{0}{0}$, so we need to use algebraic manipulation to simplify the expression first. We multiply by the conjugate to get
$$\lim _ { x ightarrow 2 } \sqrt { \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } } \cdot \frac{\sqrt{ \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } } + \sqrt{ \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } }}{\sqrt{ \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } } + \sqrt{ \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } } }$$
Simplifying, we get
$$\lim _ { x ightarrow 2 } \frac{4x^2+1}{3x-2} \cdot \frac{1}{\sqrt { \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } } + \sqrt{ \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } }}$$
$$=\frac{\lim_{xightarrow 2}(4x^2+1)}{\lim_{xightarrow 2}(3x-2)} \cdot \frac{1}{\lim _ { x ightarrow 2 } \left(\sqrt { \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } } + \sqrt{ \frac { 4 x ^ { 2 } + 1 } { 3 x - 2 } }ight)}$$
Using direct substitution, we can find the three limits in that expression. The numerator evaluates to $4(2)^2+1=17$, the denominator to $3(2)-2=4$, and the term inside the parenthesis evaluates to $2\sqrt{\frac{17}{4}} = \sqrt{17}$. So, simplifying our expression, we get
$$\frac{17}{4\sqrt{17}}=\frac{\sqrt{17}}{4}$$
Thus, the answer is $\boxed{	ext{(C)}}$.

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