evaluate the limit using lhospitals rule if necessary\n$$\\lim_{x \\to 0} \\frac{\\sin(4x)}{\\sin(13x)}$$\n
evaluate the limit using lhospitals rule if necessary\n$$\\lim_{x \\to 0} \\frac{\\sin(4x)}{\\sin(13x)}$$\n
Answer
Explanation:
Step1: Check indeterminate form
Substitute (x = 0) into (\frac{\sin(4x)}{\sin(13x)}). We get (\frac{\sin(0)}{\sin(0)}=\frac{0}{0}), which is an indeterminate form. So, L'Hospital's rule can be applied.
Step2: Apply L'Hospital's rule
Differentiate the numerator and denominator. The derivative of (\sin(4x)) with respect to (x) is (4\cos(4x)) (using the chain - rule (y=\sin(u)), (u = 4x), (y^\prime=\cos(u)\cdot u^\prime)), and the derivative of (\sin(13x)) with respect to (x) is (13\cos(13x)). So, (\lim_{x\rightarrow0}\frac{\sin(4x)}{\sin(13x)}=\lim_{x\rightarrow0}\frac{4\cos(4x)}{13\cos(13x)}).
Step3: Evaluate the new limit
Substitute (x = 0) into (\frac{4\cos(4x)}{13\cos(13x)}). Since (\cos(0)=1), we have (\frac{4\cos(0)}{13\cos(0)}=\frac{4\times1}{13\times1}).
Answer:
(\frac{4}{13})