use lhôpitals rule to find the limit. lim x→−∞ x^6 - 31 / x - 7 apply lhôpitals rule so that the lim x→−∞…

Explanation:

Step1: Differentiate numerator

The derivative of $x^{6}-31$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $6x^{5}$.

Step2: Differentiate denominator

The derivative of $x - 7$ using the power - rule is $1$.

Step3: Apply L'Hopital's rule

By L'Hopital's rule, if $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is in the indeterminate form $\frac{\infty}{\infty}$ or $\frac{0}{0}$, then $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$. So $\lim_{x\rightarrow-\infty}\frac{x^{6}-31}{x - 7}=\lim_{x\rightarrow-\infty}\frac{6x^{5}}{1}$.

Answer:

$-\infty$