Question

evaluate the limit using lhospitals rule if necessary \\(\lim_{x\to 0} \frac{\sin(7x)}{\sin(2x)}\\) question help: video message instructor submit question jump to answer

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\sin(7x)=\sin(0) = 0\) and \(\sin(2x)=\sin(0)=0\). So, \(\lim_{x
ightarrow0}\frac{\sin(7x)}{\sin(2x)}\) is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hospital's Rule

L'Hospital's Rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).
Differentiate \(y = \sin(7x)\) using the chain - rule: \((\sin(7x))^{\prime}=7\cos(7x)\).
Differentiate \(y=\sin(2x)\) using the chain - rule: \((\sin(2x))^{\prime}=2\cos(2x)\).
So, \(\lim_{x
ightarrow0}\frac{\sin(7x)}{\sin(2x)}=\lim_{x
ightarrow0}\frac{7\cos(7x)}{2\cos(2x)}\).

Step3: Evaluate the new limit

Substitute \(x = 0\) into \(\frac{7\cos(7x)}{2\cos(2x)}\).
Since \(\cos(0)=1\), we have \(\frac{7\cos(0)}{2\cos(0)}=\frac{7\times1}{2\times1}\).

Answer:

\(\frac{7}{2}\)