iamMJae
  • iamMJae
How do I start with questions that look like this?:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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iamMJae
  • iamMJae
\[Construct~an~f(x)~such~that~\lim_{x \rightarrow 3} f(x)=\infty~but~\lim_{x \rightarrow 3} (x-3)f(x)=0.\]
iamMJae
  • iamMJae
I ended up with \[f(x)=\frac{ x+3 }{ (x-3)^2 }\] It's almost right except it diverges and even if it could be correct, how do I go about my "solution" because I just guessed at it and it was somehow close.
anonymous
  • anonymous
the infinite limit is an asymptote, that's why (x-3) is in the denominator. I think you should be fine if you don't square it. Multiplying by (x-3) makes a hole at (3, 0).

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More answers

ganeshie8
  • ganeshie8
isn't squaring a must for both side limits to be +infinity ?
anonymous
  • anonymous
you're right. I missed that
ganeshie8
  • ganeshie8
looks very tricky haha i still don't see how we can get 0 for the limit f(x)*(x-3)
iamMJae
  • iamMJae
Sorry, yes, I was wrong. The square makes it not diverge.
iamMJae
  • iamMJae
@ganeshie8 It's not "limit f(x)*(x-3)" but limit (x-3)f(x)... If that makes any difference.
Empty
  • Empty
Maybe it's a typo and it should really say: \[\lim_{x \rightarrow 3} (x-3)f(x)=1\]
iamMJae
  • iamMJae
@Empty It says 0 but if it was, what would be the function?
Empty
  • Empty
Oh even then I can't find a simple answer because I had thought it would have been either \[f(x) = \frac{1}{x-3} \]or \[f(x) = \frac{1}{|x-3|} \] but both would not work either. Weird, I will be amazed if there is a solution to this question! :D
ganeshie8
  • ganeshie8
Below function fits the given spec but im pretty sure this is not what you want \[f(x)=\dfrac{1}{\sqrt{x-3}}\]
ganeshie8
  • ganeshie8
f:R->R so that we don't have to wry about x<=3 part that is not part of the domain of f(x)
amoodarya
  • amoodarya
\[\frac{1}{\sqrt[3]{x-3}}\]
amoodarya
  • amoodarya
what this ?
Empty
  • Empty
I don't think any of these work exactly, unfortunately because of two-sided limits not existing.
ganeshie8
  • ganeshie8
two sided limit is not the definition of a limit for boundary points
ganeshie8
  • ganeshie8
3 is a boundary point for f(x) = 1/sqrt(x-3), so we are fine
Empty
  • Empty
I guessss...
amoodarya
  • amoodarya
\[\frac{1}{\sqrt[3]{|x-3|}}\\ \lim_{x \rightarrow 3}\frac{1}{\sqrt[3]{|x-3|}}=+\infty\\lim_{x \rightarrow 3}(x-3)\frac{1}{\sqrt[3]{|x-3|}}=0\]
ganeshie8
  • ganeshie8
wow! thats really clever!!
amoodarya
  • amoodarya
\[f(x)=-\ln|x-3|\] this function also work
ganeshie8
  • ganeshie8
both work like charm! xD
Empty
  • Empty
wowwowowow
amoodarya
  • amoodarya
I think ,we can't find f(x) R-->R easily !
phi
  • phi
in general, if you have a function f(x) -> infinity then let g(x)= 1/f(x) goes to zero i.e. let f(x)= 1/g(x) next, (x-3) f(x) = (x-3)/g(x) goes to 0 for this to be true, we want g(x) to "go to zero *slower* * than (x-3) goes to zero. in which case (x-3) "wins"
phi
  • phi
as amood suggests, a log function grows very slowly we could also use sqrt, which grows more slowly than (x-3) for example \[ \lim_{x\rightarrow 3}\frac{1}{\sqrt{|x-3|}} = \infty \\ \lim_{x\rightarrow 3}\frac{(x-3)}{\sqrt{|x-3|}} =0 \]

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