2. (2 points) consider the functions $f(x)=999x^{999}$ and $g(x)=(1.001)^x$.\na. which function dominates as…

2. (2 points) consider the functions $f(x)=999x^{999}$ and $g(x)=(1.001)^x$.\na. which function dominates as $x\rightarrowinfty$?\nb. use your answer in (a) and find the end - behavior of the quotients:\ni. $lim_{x\rightarrowinfty}\frac{999x^{999}}{(1.001)^x}=$\nii. $lim_{x\rightarrowinfty}\frac{(1.001)^x}{999x^{999}}=$

2. (2 points) consider the functions $f(x)=999x^{999}$ and $g(x)=(1.001)^x$.\na. which function dominates as $x\rightarrowinfty$?\nb. use your answer in (a) and find the end - behavior of the quotients:\ni. $lim_{x\rightarrowinfty}\frac{999x^{999}}{(1.001)^x}=$\nii. $lim_{x\rightarrowinfty}\frac{(1.001)^x}{999x^{999}}=$

Answer

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)