Question

evaluate the limit using lhospitals rule \\(\lim\limits_{x\to 0} \frac{e^x - 1}{\sin(12x)}\\) question help: video message instructor submit question jump to answer

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\frac{e^{0}-1}{\sin(12\times0)}=\frac{1 - 1}{0}=\frac{0}{0}\). So L'Hospital's rule can be applied. L'Hospital's rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Step2: Differentiate the numerator and denominator

The derivative of \(y = e^{x}-1\) is \(y^{\prime}=e^{x}\) (using the rule \(\frac{d}{dx}(e^{x})=e^{x}\) and \(\frac{d}{dx}(c)=0\) for a constant \(c\)).
The derivative of \(y=\sin(12x)\) is \(y^{\prime}=12\cos(12x)\) (using the chain - rule \(\frac{d}{dx}(\sin(u))=\cos(u)\cdot u^{\prime}\), where \(u = 12x\) and \(u^{\prime}=12\))

Step3: Evaluate the new limit

Now we find \(\lim_{x
ightarrow0}\frac{e^{x}}{12\cos(12x)}\)
Substitute \(x = 0\) into \(\frac{e^{x}}{12\cos(12x)}\), we get \(\frac{e^{0}}{12\cos(0)}\)
Since \(e^{0}=1\) and \(\cos(0)=1\)

Answer:

\(\frac{1}{12}\)