Find the limit. $\lim _ { x \rightarrow \infty } \frac { - 6 x } { \sqrt { 64 x ^ { 2 } - 5 } }$
A) $- \frac { 3 } { 32 }$
B) $- \frac { 3 } { 4 }$
C) 1
D) $- 6$
E) $- \infty$

Question

Quizplus · Accepted Answer

To find the limit as $x$ approaches infinity of the given function, we divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x$. This simplifies the expression to $\lim _ { x \rightarrow \infty } \frac { - 6 } { \sqrt { 64 - \frac{5}{x^2} } }$. As $x$ approaches infinity, $\frac{5}{x^2}$ approaches 0, leaving us with $\frac{-6}{\sqrt{64}} = \frac{-6}{8} = -\frac{3}{4}$.

Find the Limit lim⁡x→∞−6x64x2−5\lim _ { x \rightarrow \infty } \frac { - 6 x } { \sqrt { 64 x ^ { 2 } - 5 } }limx→∞​64x2−5​−6x​

Find the Limit $\lim _ { x \rightarrow \infty } \frac { - 6 x } { \sqrt { 64 x ^ { 2 } - 5 } }$