Question

b. (2 points) \lim_{x \to 0} \frac{\sin(2x)}{3x}
c. (2 points) \lim_{x \to 0} \frac{e^{x} - 1}{\sin(x)}

Explanation:

Step1: Apply the limit property

We know that \(\lim_{u
ightarrow0}\frac{\sin u}{u} = 1\). For \(\lim_{x
ightarrow0}\frac{\sin(2x)}{3x}\), rewrite it as \(\frac{2}{3}\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}\). Let \(u = 2x\), when \(x
ightarrow0\), \(u
ightarrow0\).

Step2: Calculate the limit

Since \(\lim_{u
ightarrow0}\frac{\sin u}{u}=1\), then \(\frac{2}{3}\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}=\frac{2}{3}\times1=\frac{2}{3}\).

Step3: For \(\lim_{x

ightarrow0}\frac{e^{x}-1}{\sin(x)}\)
Use L - H rule (since \(\lim_{x
ightarrow0}(e^{x}-1)=0\) and \(\lim_{x
ightarrow0}\sin(x) = 0\)). Differentiate numerator and denominator. The derivative of \(y = e^{x}-1\) is \(y^\prime=e^{x}\), and the derivative of \(y=\sin(x)\) is \(y^\prime=\cos(x)\).

Step4: Calculate the new limit

\(\lim_{x
ightarrow0}\frac{e^{x}}{\cos(x)}\), substitute \(x = 0\) into \(\frac{e^{x}}{\cos(x)}\), we get \(\frac{e^{0}}{\cos(0)}=\frac{1}{1}=1\).

Answer:

For \(b\), the limit is \(\frac{2}{3}\). For \(c\), the limit is \(1\).