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

Question

Quizplus · Accepted Answer

Both numerator and denominator have the highest power of x as x^3. Dividing both by x^3, we get:
$$\frac{4x^3+3x^2}{x-6x^2}= \frac{4+\frac{3}{x}}{\frac{1}{x^2}-6}$$
As x tends to infinity, the fraction in the numerator tends to 4 and the denominator tends to 0. Thus, the limit is infinity.

Find the Limit lim⁡x→∞4x3+3x2x−6x2\lim _ { x \rightarrow \infty } \frac { 4 x ^ { 3 } + 3 x ^ { 2 } } { x - 6 x ^ { 2 } }x→∞lim​x−6x24x3+3x2​

Find the Limit $\lim _ { x \rightarrow \infty } \frac { 4 x ^ { 3 } + 3 x ^ { 2 } } { x - 6 x ^ { 2 } }$