Divide numerator and denominator by the highest power of x in the denominator to find the limit.
-$\lim _ { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } } { 3 + 9 x ^ { 2 } } }$
A) $\frac { 4 } { 3 }$
B) $\frac { 16 } { 9 }$
C) does not exist
D) \(\frac { 16 } { 3 }\

Question

Divide numerator and denominator by the highest power of x in the denominator to find the limit.
-$\lim _  { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } } { 3 + 9 x ^ { 2 } } }$
A) $\frac { 4 } { 3 }$
B) $\frac { 16 } { 9 }$
C) does not exist
D) \(\frac { 16 } { 3 }\

Quizplus · Accepted Answer

Divide numerator and denominator by $x^2$.$$\lim _  { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } } { 3 + 9 x ^ { 2 } } } = \lim _  { x \rightarrow \infty } \sqrt { \frac { 16 } { 3 + 9 x ^ { 2 } / x ^ { 2 } } } = \lim _  { x \rightarrow \infty } \sqrt { \frac { 16 } { 3 / x ^ { 2 } + 9 } } = \sqrt { \frac { 16 } { 0 + 9 } } = \frac { 4 } { 3 }$$

Divide Numerator and Denominator by the Highest Power of X lim⁡x→∞16x23+9x2\lim _ { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } } { 3 + 9 x ^ { 2 } } }limx→∞​3+9x216x2​​

Divide Numerator and Denominator by the Highest Power of X $\lim _ { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } } { 3 + 9 x ^ { 2 } } }$