Find (A) the leading term of the polynomial, (8) the limit as $$ x $$ approaches $$ \infty $$, and (C) the limit as $$ x $$ approaches - $$ \infty $$. $$ p(x)=19+3 x^{2}-6 x^{3} $$ (A) The loading term is $$ -6 x^{3} $$. (B) The limit of $$ p(x) $$ as $$ x $$ approaches $$ \infty $$ is $$ \square $$.

Question

UpStudy AI · Accepted Answer

Let's analyze the polynomial $$ p(x) = 19 + 3x^{2} - 6x^{3} $$ to find the required limits.

### (A) Leading Term
You correctly identified the leading term:
$$ \text{Leading term} = -6x^{3} $$

### (B) Limit as $$ x $$ Approaches $$ \infty $$
To find the limit of $$ p(x) $$ as $$ x $$ approaches $$ \infty $$, we focus on the leading term since it dominates the behavior of the polynomial for large values of $$ x $$.

$$ p(x) = -6x^{3} + 3x^{2} + 19 $$

As $$ x $$ approaches $$ \infty $$:
- The term $$ -6x^{3} $$ grows without bound in the negative direction.
- The other terms $$ 3x^{2} $$ and $$ 19 $$ become insignificant compared to $$ -6x^{3} $$.

Therefore:
$$ \lim_{x \to \infty} p(x) = -\infty $$

### (C) Limit as $$ x $$ Approaches $$ -\infty $$
Similarly, for $$ x $$ approaching $$ -\infty $$:

$$ p(x) = -6x^{3} + 3x^{2} + 19 $$

As $$ x $$ approaches $$ -\infty $$:
- The term $$ -6x^{3} $$ behaves like $$ -6(-\infty)^{3} = -6(-\infty) = +\infty $$.
- The other terms $$ 3x^{2} $$ and $$ 19 $$ remain insignificant compared to $$ -6x^{3} $$.

Therefore:
$$ \lim_{x \to -\infty} p(x) = \infty $$

### Summary of Limits
- **(B)** $$ \lim_{x \to \infty} p(x) = -\infty $$
- **(C)** $$ \lim_{x \to -\infty} p(x) = \infty $$
(A) The leading term is $$ -6x^{3} $$. (B) The limit of $$ p(x) $$ as $$ x $$ approaches $$ \infty $$ is $$ -\infty $$. (C) The limit of $$ p(x) $$ as $$ x $$ approaches $$ -\infty $$ is $$ \infty $$.

Real Tutor Solution

Answer

Solution

Extra Insights

Related Questions

Latest Calculus Questions