evaluate the following limit using lhospitals rule where appropriate.\n\n$$\\lim_{x \\to 0}…

evaluate the following limit using lhospitals rule where appropriate.\n\n$$\\lim_{x \\to 0} \\frac{\\sin(8x)}{\\tan(13x)}$$\n\nanswer:

evaluate the following limit using lhospitals rule where appropriate.\n\n$$\\lim_{x \\to 0} \\frac{\\sin(8x)}{\\tan(13x)}$$\n\nanswer:

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\rightarrow0}\frac{\sin(8x)}{\tan(13x)}=\lim_{x\rightarrow0}\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})