give a limit expression that describes the left end behavior of the function.\nf(x)=\frac{41x^{7}+4x^{2}}{18x…

give a limit expression that describes the left end behavior of the function.\nf(x)=\frac{41x^{7}+4x^{2}}{18x^{6}-2x}\n(lim_{x\rightarrow-infty}f(x)=square) (simplify your answer.)

give a limit expression that describes the left end behavior of the function.\nf(x)=\frac{41x^{7}+4x^{2}}{18x^{6}-2x}\n(lim_{x\rightarrow-infty}f(x)=square) (simplify your answer.)

Answer

Explanation:

Step1: Identify leading - terms

For the numerator $41x^{7}+4x^{2}$, the leading - term is $41x^{7}$. For the denominator $18x^{6}-2x$, the leading - term is $18x^{6}$.

Step2: Consider the limit as $x\to-\infty$

We find $\lim_{x\to-\infty}\frac{41x^{7}+4x^{2}}{18x^{6}-2x}$. Divide both the numerator and denominator by $x^{6}$ (the highest power of $x$ in the denominator). We get $\lim_{x\to-\infty}\frac{41x + \frac{4}{x^{4}}}{18-\frac{2}{x^{5}}}$.

Step3: Evaluate the limit

As $x\to-\infty$, $\frac{4}{x^{4}}\to0$ and $\frac{2}{x^{5}}\to0$. So, $\lim_{x\to-\infty}\frac{41x+\frac{4}{x^{4}}}{18 - \frac{2}{x^{5}}}=\lim_{x\to-\infty}\frac{41x}{18}=-\infty$.

Answer:

$-\infty$