Question

evaluate \\( \lim _ { x \
ightarrow 0 } \frac { 1 - e ^ { x } } { 3 x } \\) a. 0 b. \\( \infty \\) c. \\( - \frac { 1 } { 3 } \\) d. \\( \frac { 1 } { 3 } \\) e. dne answer: type your answer... 2 fill in the blank 1 point \\( \lim _ { x \
ightarrow \infty } \frac { \ln x } { x ^ { 2 } } \\) a. 0 b. \\( \infty \\) c. \\( - \infty \\) d. 1 e. dne answer: type your answer...

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, \(\lim_{x
ightarrow0}\frac{1 - e^{x}}{3x}=\lim_{x
ightarrow0}\frac{-e^{x}}{3}\)

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

Step4: Evaluate the limit

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

Answer:

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