Question

Use the limit theorem and the properties of limits to find the limit.
Picture below

Answer 1

We have

$\dfrac{-6x^3+7x+7}{8x^3-8x+5}=\dfrac{-6+\frac7{x^2}+\frac7{x^3}}{8-\frac8{x^2}+\frac5{x^3}}$

As $x\to-\infty$, all of terms with powers of $x$ in their denominators will converge to 0, leaving you with

$\displaystyle\lim_{x\to-\infty}\frac{-6x^3+7x+7}{8x^3-8x+5}=\lim_{x\to-\infty}\frac{-6}8=-\frac34$

making the answer B.

Answer 2

Answer:

The value of limit is $\dfrac{-3}{4}$

Option B is correct.  

Step-by-step explanation:

Given:

$L=\lim_{x\rightarrow -\infty}\left ( \dfrac{-6x^3+7x+7}{8x^3-8x+5}\right )$

Here we have rational function whose limit is minus infinity.

Divide numerator and denominator by highest degree of polynomial (x³)

$L=\lim_{x\rightarrow -\infty}\left ( \dfrac{-6+7/x^2+7/x^3}{8-8/x^2+5/x^3}\right )$

Apply limit and we get

$L=\left ( \dfrac{-6+\frac{7}{\infty}+\frac{7}{-\infty}}{8-\frac{8}{\infty}+\frac{5}{-\infty}}\right )$

$L=\left ( \dfrac{-6+0+0}{8-0+0}\right )$

$L=\dfrac{-6}{8}\Rightarrow -\dfrac{3}{4}$

Hence, The value of limit is $\dfrac{-3}{4}$