4.8 lhopitals rule - l5: problem 2 (6 points) evaluate the following limit using lhospitals rule where…

Explanation:

Step1: Check the form of the limit

As $x\rightarrow0$, $\sin(5x)\rightarrow0$ and $\tan(9x)\rightarrow0$. So, it is in the $\frac{0}{0}$ - form, and L'Hopital's rule can be applied.

Step2: Differentiate the numerator and denominator

The derivative of $y = \sin(5x)$ using the chain - rule is $y^\prime=5\cos(5x)$. The derivative of $y=\tan(9x)$ using the chain - rule is $y^\prime = 9\sec^{2}(9x)$. So, $\lim_{x\rightarrow0}\frac{\sin(5x)}{\tan(9x)}=\lim_{x\rightarrow0}\frac{5\cos(5x)}{9\sec^{2}(9x)}$.

Step3: Evaluate the new limit

As $x\rightarrow0$, $\cos(5x)\rightarrow1$ and $\sec^{2}(9x)\rightarrow1$. Then $\lim_{x\rightarrow0}\frac{5\cos(5x)}{9\sec^{2}(9x)}=\frac{5\times1}{9\times1}=\frac{5}{9}$.

Answer:

$\frac{5}{9}$