Question

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. lim sin(7x) csc(9x) x→0

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 \(\lim_{x
ightarrow0}\frac{\sin(7x)}{\sin(9x)}\) is in the \(\frac{0}{0}\) form (because \(\sin(0) = 0\)), by L'Hospital's Rule \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\) if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) 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, \(\lim_{x
ightarrow0}\frac{\sin(7x)}{\sin(9x)}=\lim_{x
ightarrow0}\frac{7\cos(7x)}{9\cos(9x)}\).

Step3: Evaluate the limit

Substitute \(x = 0\) into \(\frac{7\cos(7x)}{9\cos(9x)}\).
We know that \(\cos(0)=1\). So, \(\lim_{x
ightarrow0}\frac{7\cos(7x)}{9\cos(9x)}=\frac{7\cos(0)}{9\cos(0)}=\frac{7\times1}{9\times1}\).

Answer:

\(\frac{7}{9}\)