evaluate the following limit using lhospitals rule where appropriate. limₓ→0 sin(4x)/tan(15x)

Explanation:

Step1: Check the form of the limit

When (x\rightarrow0), (\sin(4x)\rightarrow0) and (\tan(15x)\rightarrow0). So, it is in the (\frac{0}{0}) form, and we can apply L'Hospital's rule. By L'Hospital's rule, if (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}) is in the (\frac{0}{0}) or (\frac{\infty}{\infty}) form, then (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}) Here, (f(x)=\sin(4x)), (f^{\prime}(x) = 4\cos(4x)) (using the chain rule ((\sin(u))^{\prime}=\cos(u)\cdot u^{\prime}), where (u = 4x) and (u^{\prime}=4)) (g(x)=\tan(15x)), (g^{\prime}(x)=15\sec^{2}(15x)) (using the chain rule ((\tan(u))^{\prime}=\sec^{2}(u)\cdot u^{\prime}), where (u = 15x) and (u^{\prime}=15))

Step2: Calculate the new limit

(\lim_{x\rightarrow0}\frac{\sin(4x)}{\tan(15x)}=\lim_{x\rightarrow0}\frac{4\cos(4x)}{15\sec^{2}(15x)}) Since (\cos(0) = 1) and (\sec(0)=\frac{1}{\cos(0)} = 1) Substitute (x = 0) into (\frac{4\cos(4x)}{15\sec^{2}(15x)}), we get (\frac{4\cos(0)}{15\sec^{2}(0)}=\frac{4\times1}{15\times1})

Answer:

(\frac{4}{15})