Question

(7)

$$ lim _ { x ightarrow 0 } \frac { e ^ { x } - 1 } { 3 x } $$

Explanation:

Step1: Apply L'Hopital's Rule

When \(x
ightarrow0\), \(\frac{e^{x}-1}{3x}\) is in the \(\frac{0}{0}\) form. According to L'Hopital'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 \(g(x) = 3x\).
\(f^{\prime}(x)=e^{x}\), \(g^{\prime}(x)=3\).
So, \(\lim_{x
ightarrow0}\frac{e^{x}-1}{3x}=\lim_{x
ightarrow0}\frac{e^{x}}{3}\).

Step2: Substitute \(x = 0\)

Substitute \(x = 0\) into \(\frac{e^{x}}{3}\).
When \(x = 0\), \(e^{0}=1\), then \(\frac{e^{0}}{3}=\frac{1}{3}\).

Answer:

\(\frac{1}{3}\)