11. -/1 points details my notes sessca find the limit. use lhospitals rule if appropriate. if there is lim x…

Explanation:

Step1: Rewrite the form

Let $t=\frac{1}{x}$, as $x\rightarrow\infty$, then $t\rightarrow0$. The limit $\lim_{x\rightarrow\infty}x\tan(\frac{8}{x})$ becomes $\lim_{t\rightarrow0}\frac{\tan(8t)}{t}$.

Step2: Apply L'Hopital's Rule

The limit $\lim_{t\rightarrow0}\frac{\tan(8t)}{t}$ is in the $\frac{0}{0}$ - form. By L'Hopital's Rule, if $\lim_{t\rightarrow a}\frac{f(t)}{g(t)}$ is in the $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then $\lim_{t\rightarrow a}\frac{f(t)}{g(t)}=\lim_{t\rightarrow a}\frac{f^{\prime}(t)}{g^{\prime}(t)}$. Here, $f(t)=\tan(8t)$ and $g(t)=t$. The derivative of $f(t)$ is $f^{\prime}(t)=8\sec^{2}(8t)$ and the derivative of $g(t)$ is $g^{\prime}(t) = 1$. So, $\lim_{t\rightarrow0}\frac{\tan(8t)}{t}=\lim_{t\rightarrow0}\frac{8\sec^{2}(8t)}{1}$.

Step3: Evaluate the limit

Substitute $t = 0$ into $\frac{8\sec^{2}(8t)}{1}$. Since $\sec(0)=1$, then $\lim_{t\rightarrow0}\frac{8\sec^{2}(8t)}{1}=8\times1^{2}=8$.

Answer:

$8$