Divide numerator and denominator by the highest power of x in the denominator to find the limit.
-$$\lim _ { x \rightarrow \infty } \frac { \sqrt [ 3 ] { x } - 6 x - 4 } { - 7 x + x ^ { 2 / 3 } + 4 }$$
A) $*$
B) 0
C) $\frac { 7 } { 6 }$
D) $\frac { 6 } { 7 }$

Question

Quizplus · Accepted Answer

Divide the numerator and denominator by the highest power of x in the denominator, which is x^(2/3). We get:
$$\lim _ { x \rightarrow \infty } \frac { \frac {1}{x^{2/3}} (\sqrt [ 3 ] { x } - 6 x - 4) } { \frac {1}{x^{2/3}}(- 7 x + x ^ { 2 / 3 } + 4)}$$
$$\lim _ { x \rightarrow \infty } \frac { \sqrt [ 3 ] { 1/x^2 } - 6/x^{5/3} - 4/x^{2/3} } { - 7/x^{1/3} + 1 + 4/x^{2/3}}$$
As x approaches infinity, all terms with a negative power of x tend to zero. Thus, we get:
$$\frac{-0-0-0}{-0+1+0} = \frac{0}{1} = 0$$
Therefore, the limit is 0, and the answer is D.

Divide Numerator and Denominator by the Highest Power of X lim⁡x→∞x3−6x−4−7x+x2/3+4\lim _ { x \rightarrow \infty } \frac { \sqrt [ 3 ] { x } - 6 x - 4 } { - 7 x + x ^ { 2 / 3 } + 4 }x→∞lim​−7x+x2/3+43x​−6x−4​

Divide Numerator and Denominator by the Highest Power of X $\lim _ { x \rightarrow \infty } \frac { \sqrt [ 3 ] { x } - 6 x - 4 } { - 7 x + x ^ { 2 / 3 } + 4 }$