evaluate the limit using lhospitals rule if necessary lim x→0 sin(13x)/sin(9x)

Explanation:

Step1: Check indeterminate form

As $x\rightarrow0$, $\sin(13x)\rightarrow0$ and $\sin(9x)\rightarrow0$. So, it's a $\frac{0}{0}$ - form.

Step2: Apply L'Hopital's rule

Differentiate the numerator and denominator. The derivative of $\sin(13x)$ with respect to $x$ is $13\cos(13x)$ and the derivative of $\sin(9x)$ with respect to $x$ is $9\cos(9x)$. So, $\lim_{x\rightarrow0}\frac{\sin(13x)}{\sin(9x)}=\lim_{x\rightarrow0}\frac{13\cos(13x)}{9\cos(9x)}$.

Step3: Evaluate the new limit

Substitute $x = 0$ into $\frac{13\cos(13x)}{9\cos(9x)}$. We know that $\cos(0)=1$. So, $\lim_{x\rightarrow0}\frac{13\cos(13x)}{9\cos(9x)}=\frac{13\cos(0)}{9\cos(0)}=\frac{13}{9}$.

Answer:

$\frac{13}{9}$