(7) $$ lim _ { x \rightarrow 0 } \frac { e ^ { x } - 1 } { 3 x } $$
(7) $$ lim _ { x \rightarrow 0 } \frac { e ^ { x } - 1 } { 3 x } $$
Answer
Explanation:
Step1: Apply L'Hopital's Rule
When (x\rightarrow0), (\frac{e^{x}-1}{3x}) is in the (\frac{0}{0}) form. According to L'Hopital's Rule, if (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}) is in (\frac{0}{0}) or (\frac{\infty}{\infty}) form, then (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 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\rightarrow0}\frac{e^{x}-1}{3x}=\lim_{x\rightarrow0}\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})