evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim x→∞ x³/7 (6/x…

Explanation:

Step1: Let $t=\frac{1}{x}$

As $x\rightarrow\infty$, then $t\rightarrow0$. The limit $\lim_{x\rightarrow\infty}\frac{x^{3}}{7}(\frac{6}{x}-\sin\frac{6}{x})$ becomes $\lim_{t\rightarrow0}\frac{1}{7t^{3}}(6t - \sin(6t))$.

Step2: Check the form

When $t = 0$, the expression $\frac{6t-\sin(6t)}{7t^{3}}$ is in the $\frac{0}{0}$ - form. So, we can apply L'Hopital's Rule. Differentiate the numerator and denominator. The derivative of $6t-\sin(6t)$ is $6 - 6\cos(6t)$, and the derivative of $7t^{3}$ is $21t^{2}$. So the limit becomes $\lim_{t\rightarrow0}\frac{6 - 6\cos(6t)}{21t^{2}}$.

Step3: Check the form again

When $t = 0$, $\frac{6 - 6\cos(6t)}{21t^{2}}$ is still in the $\frac{0}{0}$ - form. Apply L'Hopital's Rule again. Differentiate the numerator and denominator. The derivative of $6 - 6\cos(6t)$ is $36\sin(6t)$, and the derivative of $21t^{2}$ is $42t$. So the limit becomes $\lim_{t\rightarrow0}\frac{36\sin(6t)}{42t}$.

Step4: Check the form again

When $t = 0$, $\frac{36\sin(6t)}{42t}$ is in the $\frac{0}{0}$ - form. Apply L'Hopital's Rule again. Differentiate the numerator and denominator. The derivative of $36\sin(6t)$ is $216\cos(6t)$, and the derivative of $42t$ is $42$.

Step5: Evaluate the limit

Now, $\lim_{t\rightarrow0}\frac{216\cos(6t)}{42}=\frac{216\cos(0)}{42}=\frac{216}{42}=\frac{36}{7}$.

Answer:

$\frac{36}{7}$