find the following limit. if the limit is infinite, enter infinity or -infinity. if the limit is not…

Explanation:

Step1: Divide numerator and denominator by highest - power of x

Divide both the numerator and denominator of the function $\frac{x^{4}-3x^{3}-4}{5 - 5x^{2}+3x^{4}}$ by $x^{4}$. We get $\lim_{x\rightarrow-\infty}\frac{\frac{x^{4}}{x^{4}}-\frac{3x^{3}}{x^{4}}-\frac{4}{x^{4}}}{\frac{5}{x^{4}}-\frac{5x^{2}}{x^{4}}+\frac{3x^{4}}{x^{4}}}=\lim_{x\rightarrow-\infty}\frac{1-\frac{3}{x}-\frac{4}{x^{4}}}{\frac{5}{x^{4}}-\frac{5}{x^{2}} + 3}$.

Step2: Evaluate the limit of each term

As $x\rightarrow-\infty$, we know that $\lim_{x\rightarrow-\infty}\frac{1}{x^{n}} = 0$ for $n>0$. So, $\lim_{x\rightarrow-\infty}\frac{3}{x}=0$, $\lim_{x\rightarrow-\infty}\frac{4}{x^{4}} = 0$, $\lim_{x\rightarrow-\infty}\frac{5}{x^{4}}=0$ and $\lim_{x\rightarrow-\infty}\frac{5}{x^{2}}=0$.

Step3: Calculate the limit of the function

Substitute the limits of the individual terms into the function: $\frac{\lim_{x\rightarrow-\infty}(1)-\lim_{x\rightarrow-\infty}\frac{3}{x}-\lim_{x\rightarrow-\infty}\frac{4}{x^{4}}}{\lim_{x\rightarrow-\infty}\frac{5}{x^{4}}-\lim_{x\rightarrow-\infty}\frac{5}{x^{2}}+\lim_{x\rightarrow-\infty}3}=\frac{1 - 0-0}{0 - 0+3}=\frac{1}{3}$.

Answer:

$\frac{1}{3}$