## iamMJae one year ago How do I start with questions that look like this?:

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1. iamMJae

$Construct~an~f(x)~such~that~\lim_{x \rightarrow 3} f(x)=\infty~but~\lim_{x \rightarrow 3} (x-3)f(x)=0.$

2. 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.

3. 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).

4. ganeshie8

isn't squaring a must for both side limits to be +infinity ?

5. anonymous

you're right. I missed that

6. ganeshie8

looks very tricky haha i still don't see how we can get 0 for the limit f(x)*(x-3)

7. iamMJae

Sorry, yes, I was wrong. The square makes it not diverge.

8. iamMJae

@ganeshie8 It's not "limit f(x)*(x-3)" but limit (x-3)f(x)... If that makes any difference.

9. Empty

Maybe it's a typo and it should really say: $\lim_{x \rightarrow 3} (x-3)f(x)=1$

10. iamMJae

@Empty It says 0 but if it was, what would be the function?

11. 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

12. ganeshie8

Below function fits the given spec but im pretty sure this is not what you want $f(x)=\dfrac{1}{\sqrt{x-3}}$

13. 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)

14. amoodarya

$\frac{1}{\sqrt[3]{x-3}}$

15. amoodarya

what this ?

16. Empty

I don't think any of these work exactly, unfortunately because of two-sided limits not existing.

17. ganeshie8

two sided limit is not the definition of a limit for boundary points

18. ganeshie8

3 is a boundary point for f(x) = 1/sqrt(x-3), so we are fine

19. Empty

I guessss...

20. amoodarya

$\frac{1}{\sqrt[3]{|x-3|}}\\ \lim_{x \rightarrow 3}\frac{1}{\sqrt[3]{|x-3|}}=+\infty\\lim_{x \rightarrow 3}(x-3)\frac{1}{\sqrt[3]{|x-3|}}=0$

21. ganeshie8

wow! thats really clever!!

22. amoodarya

$f(x)=-\ln|x-3|$ this function also work

23. ganeshie8

both work like charm! xD

24. Empty

wowwowowow

25. amoodarya

I think ,we can't find f(x) R-->R easily !

26. 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"

27. phi

as amood suggests, a log function grows very slowly we could also use sqrt, which grows more slowly than (x-3) for example $\lim_{x\rightarrow 3}\frac{1}{\sqrt{|x-3|}} = \infty \\ \lim_{x\rightarrow 3}\frac{(x-3)}{\sqrt{|x-3|}} =0$