Question

for which of the following does $lim_{x
ightarrowinfty}f(x)=0$?
i $f(x)=\frac{ln x}{x}$
ii $f(x)=\frac{x}{ln x}$
iii $f(x)=\frac{e^{x}}{x^{2}}$
(a) i only
(b) ii only
(c) iii only

Explanation:

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

As \(x
ightarrow\infty\), use L - H rule. Differentiate numerator and denominator. The derivative of \(\ln x\) is \(\frac{1}{x}\) and of \(x\) is \(1\). So \(\lim_{x
ightarrow\infty}\frac{\ln x}{x}=\lim_{x
ightarrow\infty}\frac{\frac{1}{x}}{1}=0\).

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

As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{x}{\ln x}=\infty\) since the numerator \(x\) grows faster than the denominator \(\ln x\) as \(x\) gets large.

Step3: Analyze \(f(x)=\frac{e^{x}}{x^{2}}\)

As \(x
ightarrow\infty\), use L - H rule multiple - times. First, \(\lim_{x
ightarrow\infty}\frac{e^{x}}{x^{2}}=\lim_{x
ightarrow\infty}\frac{e^{x}}{2x}\) (by differentiating numerator and denominator). Then applying L - H rule again, \(\lim_{x
ightarrow\infty}\frac{e^{x}}{2x}=\lim_{x
ightarrow\infty}\frac{e^{x}}{2}=\infty\).

Answer:

A. I only