find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…

Explanation:

Step1: Rewrite the function

We know that (\csc(9x)=\frac{1}{\sin(9x)}). So, (\sin(7x)\csc(9x)=\frac{\sin(7x)}{\sin(9x)}).

Step2: Use L'Hospital's Rule

Since (\lim_{x\rightarrow0}\frac{\sin(7x)}{\sin(9x)}) is in the (\frac{0}{0}) form (because (\sin(0) = 0)), by L'Hospital's Rule (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}) if (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}) is in (\frac{0}{0}) or (\frac{\infty}{\infty}) form. Differentiate (y = \sin(7x)) with respect to (x): (y^{\prime}=7\cos(7x)) (using the chain - rule ((\sin(u))^{\prime}=\cos(u)\cdot u^{\prime}), where (u = 7x) and (u^{\prime}=7)). Differentiate (y=\sin(9x)) with respect to (x): (y^{\prime}=9\cos(9x)) (using the chain - rule ((\sin(u))^{\prime}=\cos(u)\cdot u^{\prime}), where (u = 9x) and (u^{\prime}=9)). So, (\lim_{x\rightarrow0}\frac{\sin(7x)}{\sin(9x)}=\lim_{x\rightarrow0}\frac{7\cos(7x)}{9\cos(9x)}).

Step3: Evaluate the limit

Substitute (x = 0) into (\frac{7\cos(7x)}{9\cos(9x)}). We know that (\cos(0)=1). So, (\lim_{x\rightarrow0}\frac{7\cos(7x)}{9\cos(9x)}=\frac{7\cos(0)}{9\cos(0)}=\frac{7\times1}{9\times1}).

Answer:

(\frac{7}{9})