Question

evaluate the limit using lhospitals rule \\(\lim_{x\to 0} \frac{\sin(7x)}{\tan(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 \(\tan(2x)=\tan(0)=0\). So, \(\lim_{x
ightarrow0}\frac{\sin(7x)}{\tan(2x)}\) is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and denominator.
The derivative of \(y = \sin(7x)\) using the chain - rule \((u = 7x,y=\sin(u))\), \(y^\prime=\cos(u)\cdot u^\prime\), so \(y^\prime=\cos(7x)\cdot7\).
The derivative of \(y=\tan(2x)\) using the chain - rule \((u = 2x,y = \tan(u))\), \(y^\prime=\sec^{2}(u)\cdot u^\prime\), so \(y^\prime=\sec^{2}(2x)\cdot2\).
By L'Hospital's Rule, \(\lim_{x
ightarrow0}\frac{\sin(7x)}{\tan(2x)}=\lim_{x
ightarrow0}\frac{7\cos(7x)}{2\sec^{2}(2x)}\).

Step3: Evaluate the new limit

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

Answer:

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