evaluate \\( \\lim _ { x \\rightarrow 0 } \\frac { 1 - e ^ { x } } { 3 x } \\) a. 0 b. \\( \\infty \\) c…

Explanation:

Step1: Apply L'Hopital's Rule for the first limit

When (x = 0), (\frac{1 - e^{x}}{3x}) is in the (\frac{0}{0}) form. By L'Hopital's Rule, (\lim_{x\rightarrow0}\frac{1 - e^{x}}{3x}=\lim_{x\rightarrow0}\frac{-e^{x}}{3})

Step2: Substitute (x = 0)

Substitute (x = 0) into (\frac{-e^{x}}{3}), we get (\frac{-e^{0}}{3}=-\frac{1}{3})

Step3: Apply L'Hopital's Rule for the second limit

When (x\rightarrow\infty), (\frac{\ln x}{x^{2}}) is in the (\frac{\infty}{\infty}) form. By L'Hopital's Rule, (\lim_{x\rightarrow\infty}\frac{\ln x}{x^{2}}=\lim_{x\rightarrow\infty}\frac{\frac{1}{x}}{2x}=\lim_{x\rightarrow\infty}\frac{1}{2x^{2}})

Step4: Evaluate the limit

As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{1}{2x^{2}} = 0)

Answer:

1. c. (-\frac{1}{3})
2. a. (0)