question 13 · 1 point evaluate the following limit using lhospitals rule. lim(x→0) (9 sin(5x))/(5 tan(4x))…

Explanation:

Step1: Recall L'Hospital's rule

If $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is in the $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$. As $x\rightarrow0$, $\sin(5x)\rightarrow0$ and $\tan(4x)\rightarrow0$, so we can apply L'Hospital's rule.

Step2: Differentiate the numerator and denominator

The derivative of $y = 9\sin(5x)$ using the chain - rule: If $y = 9\sin(u)$ and $u = 5x$, then $y'=9\cos(u)\cdot5 = 45\cos(5x)$. The derivative of $y = 5\tan(4x)$ using the chain - rule: If $y = 5\tan(u)$ and $u = 4x$, then $y'=5\sec^{2}(u)\cdot4=20\sec^{2}(4x)$. So, $\lim_{x\rightarrow0}\frac{9\sin(5x)}{5\tan(4x)}=\lim_{x\rightarrow0}\frac{45\cos(5x)}{20\sec^{2}(4x)}$.

Step3: Evaluate the new limit

Substitute $x = 0$ into $\frac{45\cos(5x)}{20\sec^{2}(4x)}$. Since $\cos(0)=1$ and $\sec(0) = 1$, we have $\frac{45\cos(0)}{20\sec^{2}(0)}=\frac{45\times1}{20\times1}=\frac{9}{4}$.

Answer:

$\frac{9}{4}$