Question

evaluate the following limit using lhospitals rule where appropriate.

$$\lim_{x \to 0} \frac{\sin(8x)}{\tan(13x)}$$

answer:

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\sin(8x)=\sin(0) = 0\) and \(\tan(13x)=\tan(0)=0\). So, it is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and denominator.
The derivative of \(y=\sin(8x)\) using the chain rule \((u = 8x,y=\sin(u))\): \(\frac{dy}{dx}=\cos(8x)\cdot8\).
The derivative of \(y = \tan(13x)\) using the chain rule \((u = 13x,y=\tan(u))\): \(\frac{dy}{dx}=\sec^{2}(13x)\cdot13\).
So, \(\lim_{x
ightarrow0}\frac{\sin(8x)}{\tan(13x)}=\lim_{x
ightarrow0}\frac{8\cos(8x)}{13\sec^{2}(13x)}\).

Step3: Evaluate the new limit

Substitute \(x = 0\) into \(\frac{8\cos(8x)}{13\sec^{2}(13x)}\).
Since \(\cos(0)=1\) and \(\sec(0)=\frac{1}{\cos(0)} = 1\), we have \(\frac{8\cos(0)}{13\sec^{2}(0)}=\frac{8\times1}{13\times1}\).

Answer:

\(\frac{8}{13}\)