Question

directions: write limit statements for the end - behavior of the following rational functions.

1. (y=\frac{6x^{3}+2x^{3}-5x + 6}{6x^{2}+10x^{2}-4x - 12}) left: (-infty) right: (infty)
2. (y=\frac{(3x - 1)(2x+5)}{(4x + 3)^{2}}) left: right:
3. (y=\frac{x^{3}}{x - 1}) left: right:

Explanation:

Step1: Analyze degree of polynomials

For rational functions $\frac{f(x)}{g(x)}$, compare degrees of $f(x)$ and $g(x)$.

Step2: Determine left - hand limit for problem 10

As $x\to-\infty$ for $y = \frac{2x^{3}-5x + 6}{6x^{3}+10x^{2}-4x - 12}$, the leading terms are $2x^{3}$ and $6x^{3}$. Since the degrees are the same (both 3), the limit is the ratio of leading - coefficients. So $\lim_{x\to-\infty}\frac{2x^{3}-5x + 6}{6x^{3}+10x^{2}-4x - 12}=\frac{2}{6}=\frac{1}{3}$.

Step3: Determine right - hand limit for problem 10

As $x\to\infty$, for the same function $y=\frac{2x^{3}-5x + 6}{6x^{3}+10x^{2}-4x - 12}$, since the degrees of the numerator and denominator are the same (3), the limit is the ratio of leading - coefficients, $\lim_{x\to\infty}\frac{2x^{3}-5x + 6}{6x^{3}+10x^{2}-4x - 12}=\frac{2}{6}=\frac{1}{3}$.

Step4: Analyze problem 11

For $y=\frac{(4x + 3)^{2}}{(3x-1)(2x + 5)}$, expand the numerator: $(4x + 3)^{2}=16x^{2}+24x + 9$, and expand the denominator: $(3x-1)(2x + 5)=6x^{2}+15x-2x - 5=6x^{2}+13x - 5$.

Step5: Determine left - hand limit for problem 11

As $x\to-\infty$, since the degrees of the numerator and denominator are the same (2), the limit is the ratio of leading - coefficients. $\lim_{x\to-\infty}\frac{16x^{2}+24x + 9}{6x^{2}+13x - 5}=\frac{16}{6}=\frac{8}{3}$.

Step6: Determine right - hand limit for problem 11

As $x\to\infty$, since the degrees of the numerator and denominator are the same (2), the limit is the ratio of leading - coefficients. $\lim_{x\to\infty}\frac{16x^{2}+24x + 9}{6x^{2}+13x - 5}=\frac{16}{6}=\frac{8}{3}$.

Step7: Analyze problem 12

For $y=\frac{x^{2}}{\frac{x^{3}}{x - 1}}=\frac{x^{2}(x - 1)}{x^{3}}=\frac{x^{3}-x^{2}}{x^{3}}=1-\frac{1}{x}$.

Step8: Determine left - hand limit for problem 12

As $x\to-\infty$, $\lim_{x\to-\infty}(1-\frac{1}{x}) = 1$.

Step9: Determine right - hand limit for problem 12

As $x\to\infty$, $\lim_{x\to\infty}(1-\frac{1}{x}) = 1$.

Answer:

1. Left: $\frac{1}{3}$, Right: $\frac{1}{3}$
2. Left: $\frac{8}{3}$, Right: $\frac{8}{3}$
3. Left: $1$, Right: $1$