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Mathematics 19 Online
OpenStudy (anonymous):

Show that the two functions grow at the same rate: sqrt(9^x + 2^x) and 3^x

OpenStudy (anonymous):

I've tried a ton to manipulate the limit, but I just can't get it to turn into a constant that isn't 0

OpenStudy (anonymous):

I tried squaring the part inside of the limit, but that didn't help all that much

OpenStudy (anonymous):

oh nvm

OpenStudy (anonymous):

This may or may not use l'hopital btw

OpenStudy (anonymous):

yeah you can use l'hopital's rule to find that the\[\lim_{x \rightarrow \infty}(\sqrt{9^x+2^x)}/(3^x)) = 1\] If you just plug in infinity for x you'll get infinity/infinity which is undefined, so you have to use l'hopital's rule ( differentiate the top and bottom until you dont get infinity/infinity)

OpenStudy (anonymous):

Except that by using l'Hopital, it overcomplicates the problem a tremendous amount

OpenStudy (anonymous):

You get all of these nasty natural logs in there, in addition to the regular exponential functions that are already in the problem, so it only makes the problem worse

OpenStudy (anonymous):

@e.mccormick If you can help me that would be greatly appreciated

OpenStudy (anonymous):

No matter what I did, I just could not get that to simplify inside of the limit :/

OpenStudy (zarkon):

you don't need to use L'Hospitals rule

OpenStudy (anonymous):

Yeah, I kinda figured because it wasn't going to help at all

OpenStudy (zarkon):

put the \(3^x\) under the square root

OpenStudy (anonymous):

Ok, so that would make it 3^2x right?

OpenStudy (zarkon):

you can write it another way

OpenStudy (anonymous):

How?

OpenStudy (amorfide):

functions grow at the same rate if \[\lim_{x \rightarrow \infty} \frac{ f(x) }{g(x) } = M \neq 0\] where M is a finite non zero number

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

I need the limit to check out to a non-zero constant

OpenStudy (amorfide):

I cheated but there you go

OpenStudy (anonymous):

There was some black magic in that solution

OpenStudy (zarkon):

you can also write it as \[\frac{\sqrt{9^x+2^x}}{3^x}=\sqrt{\frac{9^x+2^x}{9^x}}=\sqrt{1+\left(\frac{2}{9}\right)^x}\]

OpenStudy (amorfide):

where was the black magic lmao

OpenStudy (anonymous):

Whooooa

OpenStudy (anonymous):

Oh my god

OpenStudy (anonymous):

Can you just pull down the 2 from the exponent like that to square the 3?

OpenStudy (amorfide):

I feel like wafflemans brain just got blown

OpenStudy (anonymous):

Because that is some crazy stuff right there

OpenStudy (anonymous):

I just became a toasted waffle

OpenStudy (zarkon):

\[(3^x)^2=3^{2x}=(3^2)^x=9^x\]

OpenStudy (anonymous):

Mind blown level=Hiroshima

OpenStudy (amorfide):

basically he is saying \[\sqrt{9^{x}} = 3^{x}\]

OpenStudy (amorfide):

yeah what zarkon said lol

OpenStudy (anonymous):

That is crazy. Thanks guys. I never would have even considered doing that.

OpenStudy (amorfide):

I tried to simplify myself and gave up because I never would have gotta that stuff either tbh sucks to be you

OpenStudy (anonymous):

Hahaha. BC calc IS pretty miserable at times

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