Question

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

Explanation:

Step1: Check the form of the limit

When \(x
ightarrow0\), \(\sin(4x)
ightarrow0\) and \(\tan(15x)
ightarrow0\). So, it is in the \(\frac{0}{0}\) form, and we can apply L'Hospital's rule.
By L'Hospital's rule, if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)
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

\(\lim_{x
ightarrow0}\frac{\sin(4x)}{\tan(15x)}=\lim_{x
ightarrow0}\frac{4\cos(4x)}{15\sec^{2}(15x)}\)
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:

\(\frac{4}{15}\)