Question

c. (2 points) $\lim_{x \to 0} \frac{e^{x}-1}{\sin(x)}$
d. (4 points) $\lim_{x \to 0} (\frac{1}{x}-\frac{1}{\sin(x)})$
(hint: rewrite the form and apply lhôpitals rule twice)

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\frac{e^{x}-1}{\sin(x)}\) is in the \(\frac{0}{0}\) form (\(e^{0}-1=0\) and \(\sin(0) = 0\)). By L'Hôpital's rule, if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).
Differentiate \(f(x)=e^{x}-1\) and \(g(x)=\sin(x)\). \(f^{\prime}(x)=e^{x}\), \(g^{\prime}(x)=\cos(x)\).
So, \(\lim_{x
ightarrow0}\frac{e^{x}-1}{\sin(x)}=\lim_{x
ightarrow0}\frac{e^{x}}{\cos(x)}\)

Step2: Evaluate the new limit

Substitute \(x = 0\) into \(\frac{e^{x}}{\cos(x)}\). When \(x = 0\), \(e^{0}=1\) and \(\cos(0)=1\). So \(\lim_{x
ightarrow0}\frac{e^{x}}{\cos(x)}=\frac{1}{1}=1\)

Step1: Rewrite the expression

\(\lim_{x
ightarrow0}(\frac{1}{x}-\frac{1}{\sin(x)})=\lim_{x
ightarrow0}\frac{\sin(x)-x}{x\sin(x)}\)
When \(x = 0\), \(\frac{\sin(x)-x}{x\sin(x)}\) is in the \(\frac{0}{0}\) form (\(\sin(0)-0 = 0\) and \(0\times\sin(0)=0\))

Step2: Apply L'Hôpital's rule first - time

Differentiate \(f(x)=\sin(x)-x\) and \(g(x)=x\sin(x)\).
\(f^{\prime}(x)=\cos(x)-1\), \(g^{\prime}(x)=\sin(x)+x\cos(x)\)
So, \(\lim_{x
ightarrow0}\frac{\sin(x)-x}{x\sin(x)}=\lim_{x
ightarrow0}\frac{\cos(x)-1}{\sin(x)+x\cos(x)}\)
When \(x = 0\), \(\frac{\cos(0)-1}{\sin(0)+0\times\cos(0)}=\frac{1 - 1}{0+0}=\frac{0}{0}\) (still \(\frac{0}{0}\) form)

Step3: Apply L'Hôpital's rule second - time

Differentiate \(f_1(x)=\cos(x)-1\) and \(g_1(x)=\sin(x)+x\cos(x)\)
\(f_1^{\prime}(x)=-\sin(x)\), \(g_1^{\prime}(x)=\cos(x)+\cos(x)-x\sin(x)=2\cos(x)-x\sin(x)\)
So, \(\lim_{x
ightarrow0}\frac{\cos(x)-1}{\sin(x)+x\cos(x)}=\lim_{x
ightarrow0}\frac{-\sin(x)}{2\cos(x)-x\sin(x)}\)

Step4: Evaluate the limit

Substitute \(x = 0\) into \(\frac{-\sin(x)}{2\cos(x)-x\sin(x)}\). When \(x = 0\), \(-\sin(0)=0\), \(2\cos(0)-0\times\sin(0)=2\)
So, \(\lim_{x
ightarrow0}\frac{-\sin(x)}{2\cos(x)-x\sin(x)} = 0\)

Answer:

\(\lim_{x
ightarrow0}\frac{e^{x}-1}{\sin(x)} = 1\)

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