anonymous
  • anonymous
Does anyone know how to do limits? I have a few homework questions I can't figure out.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
Post them! :)
anonymous
  • anonymous
I figured out the answer to the first question. I got the answer 6.
myininaya
  • myininaya
\[\lim_{x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}\]

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myininaya
  • myininaya
oh you did the first one ok
anonymous
  • anonymous
The rest I am clueless on. I don't even know where to start.
myininaya
  • myininaya
\[\lim_{x \rightarrow 0^+}\frac{1}{\sqrt[10](x)}\] first of all the limit is not 2
myininaya
  • myininaya
plug in even closer values to the right of 0 and you will see the number is getting really freaking huge so the limit is infinity also 1/(x^(1/10)) has a vertical asymptote at x=0 so we know it goes to infinity from the right
anonymous
  • anonymous
Okay great, I was on the right track because I kept getting a large number also.
myininaya
  • myininaya
the 2nd one is ln(9) because i remember \[f'(9)=\frac{1}{ \ln(a)} \lim_{x \rightarrow 0}\frac{9^x-9^0}{x-0}=\ln(9)\] where \[f(x)=\frac{a^x}{\ln(a)} \] But i'm not sure if you know this yet
myininaya
  • myininaya
\[f'(x)=\frac{1}{ \ln(a)} \lim_{x \rightarrow 0}\frac{9^x-9^0}{x-0}=\ln(9)*\]
myininaya
  • myininaya
and where a=9
anonymous
  • anonymous
no I have not learned that yet.
myininaya
  • myininaya
ok then just plug in values really close to 0 on both sides and see if it gets closer to anything
anonymous
  • anonymous
I am getting close to 2. Am I way off? I am not sure if I am calculating correct.
myininaya
  • myininaya
ln(9) is close to 2
myininaya
  • myininaya
that is pretty close
myininaya
  • myininaya
2.2 is a better number
anonymous
  • anonymous
yes, I had 2.1972. We have only had whole number answers in our homework so far.
myininaya
  • myininaya
well limits don't have to be whole numbers the numerical approach is okay but it only gives you approximations so if you want to put 2 i think that is close enough
anonymous
  • anonymous
Thank you so much! I think I kind of knew what I was doing on these problems but didn't trust myself. Can you go over the 4th question with me.
myininaya
  • myininaya
I would make a piecewise function here
myininaya
  • myininaya
well you know what lets not do that
anonymous
  • anonymous
I don't know what a piecewise function is.
myininaya
  • myininaya
lets just say this \[f(x)=\frac{(x-8)(x+17)}{x-8}=\frac{x^2+17x-8x- 136}{x-8}=\frac{x^2+9x-136}{x-8}\]
myininaya
  • myininaya
f is not defined at x=8 but tell me what happens as x->8 ?
anonymous
  • anonymous
25
myininaya
  • myininaya
that's right!
myininaya
  • myininaya
you know how i came up with my function?
anonymous
  • anonymous
So would I write down the function exists? I don't understand how to answer what they are asking. And yes please explain how you came up with the function.
myininaya
  • myininaya
they want you to find a function f(x)=blah such that f(8) does not exist and as x->8, f->25
myininaya
  • myininaya
so I knew I wanted this from the f(8) doesn't exist part \[f(x)=\frac{something}{x-8}\]
myininaya
  • myininaya
but then i also since i wanted the limit to exist at x=8 so that means i wanted the x-8 on bottom to cancel out with one on top \[f(x)=\frac{(x-8)(something)}{x-8}\] but we still need a little more something on top since we don't just want 1 because we want as x->8, f->25
myininaya
  • myininaya
so since the (x-8)/(x-8)->1 as x->8 we only need that something to go to 25 as x->8
myininaya
  • myininaya
so I knew 8+17=25
myininaya
  • myininaya
if x->8 then x+17->25
myininaya
  • myininaya
since 8+17 is 25
myininaya
  • myininaya
\[f(x)=\frac{(x-8)(x+17)}{x-8}\]
anonymous
  • anonymous
I would have never figured that out. Thank you so much. And your explanation is awesome! You are one smart cookie! Too bad I couldn't have you tutor me for the res of the semester. It's very hard taking a math class online.
myininaya
  • myininaya
i bet it is just keep coming back to openstudy and i or someone else will be here to help you out
myininaya
  • myininaya
people are addicted to this site
anonymous
  • anonymous
Well you are wonderful! You made my day! Thank you again!
myininaya
  • myininaya
have a good day
myininaya
  • myininaya
and thanks and welcome lol

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