Question

help

Answer 1

for part \(a\) you may argue like this

\(\lim\limits_{x\to\infty} \dfrac{x^a}{\ln x} = \infty \) implies there exists some integer \( N\) after which the expression \(\large \dfrac{x^a}{\ln x}\) is greater than \(2\) for ALL \(x\gt N\) :
\[\dfrac{x^a}{\ln x} \gt 2 \implies x^a \gt 2 \ln x\]

Answer 2

use that to prove part \(b\) : for all \(x\gt N\) we have
\[x^a \gt 2\ln x \\ -x^a\lt -2\ln x \\-x^a\lt \ln x^{-2}\\e^{-x^a} \lt x^{-2}\]

Answer 3

use comparison test for part c

Answer 4

what would you compare it to in part c?

Answer 5

i just used the argument that it is equal to 1/e^x^a which is a p integral where p is above 1 on the integral 1 to infinity

Answer 6

you must know the p-series test : \(\int\limits_0^{\infty} x^{-2} dx\) converges

Answer 7

Ok i get it now. Thank you

Answer 8

try somehting like this for part \(c\) :

we have \(e^{-x^a} \lt x^{-2}\) and we know that \( \int \limits_0^{\infty}x^{-2}dx\) converges by p-series test

So By comparison test, \(\int\limits_0^{\infty}e^{-x^a}dx\) converges too.

Answer 9

yeah thats what i said. Thank you for your help

Answer 10

hey sry lower bound must be \(1\) i guess

Answer 11

yeah i made sure to include that in my work but thanks to you i actually saw the connection and was able to do it. I was unsure how to approach it before

Answer 12

sounds good :)

Answer 13

I have two more questions if its not too much trouble.

Answer 14

I know i should use the comparison tests for them but the way my professor explained it im not sure how to do it...

Answer 15

sure tag me, il try..