Question

give a limit expression that describes the left - end behavior of the function.
f(x)=\frac{40x^{7}+x^{2}}{17x^{6}-3x}
lim_{x
ightarrow-infty}f(x)=square \text{ (simplify your answer.)}

Explanation:

Step1: Identify highest - degree terms

For the numerator $40x^{7}+x^{2}$, the highest - degree term is $40x^{7}$. For the denominator $17x^{6}-3x$, the highest - degree term is $17x^{6}$.

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

We find $\lim_{x\to-\infty}\frac{40x^{7}+x^{2}}{17x^{6}-3x}$ by dividing both numerator and denominator by $x^{6}$ (the highest - power of $x$ in the denominator). So we have $\lim_{x\to-\infty}\frac{40x + \frac{1}{x^{4}}}{17-\frac{3}{x^{5}}}$.

Step3: Evaluate the limit of each term

As $x\to-\infty$, $\frac{1}{x^{4}}\to0$ and $\frac{3}{x^{5}}\to0$. Then $\lim_{x\to-\infty}\frac{40x+\frac{1}{x^{4}}}{17 - \frac{3}{x^{5}}}=\lim_{x\to-\infty}\frac{40x}{17}$.

Step4: Determine the final limit

Since $x\to-\infty$, $40x\to-\infty$. So $\lim_{x\to-\infty}\frac{40x}{17}=-\infty$.

Answer:

$-\infty$