Question

for which of the following does \\( \lim _ { x \
ightarrow \infty } f ( x ) = 0 \\)?

i. \\( f ( x ) = \frac { \ln x } { x ^ { 100 } } \\)

ii. \\( f ( x ) = \frac { e ^ { x } } { \ln x } \\)

iii. \\( f ( x ) = \frac { x ^ { 100 } } { e ^ { x } } \\)

a i only

b ii only

c iii only

d i and ii only

e i and iii only

Explanation:

Step1: Analyze \( f(x)=\frac{\ln x}{x^{99}} \)

Use L - H rule (\( \frac{\infty}{\infty} \) form as \( x\to\infty \), \( \ln x\to\infty \), \( x^{99}\to\infty \)).
Differentiate numerator and denominator: \( \lim_{x\to\infty}\frac{\ln x}{x^{99}}=\lim_{x\to\infty}\frac{\frac{1}{x}}{99x^{98}}=\lim_{x\to\infty}\frac{1}{99x^{99}} = 0 \).

Step2: Analyze \( f(x)=\frac{e^{x}}{\ln x} \)

As \( x\to\infty \), \( e^{x}\to\infty \) and \( \ln x\to\infty \), but \( e^{x} \) grows much faster than \( \ln x \).
\( \lim_{x\to\infty}\frac{e^{x}}{\ln x}=\infty \) (since \( y = e^{x} \) is an exponential function with a much higher growth - rate than the logarithmic function \( y=\ln x \)).

Step3: Analyze \( f(x)=\frac{x^{99}}{e^{x}} \)

Use L - H rule multiple times.
First application (\( \frac{\infty}{\infty} \) form): \( \lim_{x\to\infty}\frac{x^{99}}{e^{x}}=\lim_{x\to\infty}\frac{99x^{98}}{e^{x}} \).
After 99 applications of L - H rule: \( \lim_{x\to\infty}\frac{99!}{e^{x}}=0 \) (because \( e^{x}\to\infty \) as \( x\to\infty \)).

Answer:

E. I and III only