Question

1. (2 points) consider the functions $f(x)=999x^{999}$ and $g(x)=(1.001)^x$.

a. which function dominates as $x
ightarrowinfty$?
b. use your answer in (a) and find the end - behavior of the quotients:
i. $lim_{x
ightarrowinfty}\frac{999x^{999}}{(1.001)^x}=$
ii. $lim_{x
ightarrowinfty}\frac{(1.001)^x}{999x^{999}}=$

Explanation:

Step1: Recall growth - rate of functions

Exponential functions \(a^x\) (\(a> 1\)) grow faster than polynomial functions \(x^n\) as \(x\to\infty\). Here, \(f(x)=999x^{999}\) is a polynomial function and \(g(x)=(1.001)^x\) is an exponential function with \(a = 1.001>1\).

Step2: Determine the dominant function

Since exponential functions with base \(a>1\) dominate polynomial functions as \(x\to\infty\), the function \(g(x)=(1.001)^x\) dominates \(f(x)=999x^{999}\) as \(x\to\infty\).

Step3: Find the limit of \(\lim_{x\to\infty}\frac{999x^{999}}{(1.001)^x}\)

As \(g(x)\) dominates \(f(x)\), \(\lim_{x\to\infty}\frac{999x^{999}}{(1.001)^x}=0\).

Step4: Find the limit of \(\lim_{x\to\infty}\frac{(1.001)^x}{999x^{999}}\)

Since \(g(x)\) dominates \(f(x)\), \(\lim_{x\to\infty}\frac{(1.001)^x}{999x^{999}}=\infty\).

Answer:

a. The function \(g(x)=(1.001)^x\) dominates as \(x\to\infty\).
b.
i. \(\lim_{x\to\infty}\frac{999x^{999}}{(1.001)^x}=0\)
ii. \(\lim_{x\to\infty}\frac{(1.001)^x}{999x^{999}}=\infty\)