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

you're right. I missed that

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

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

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

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

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)

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

what this ?

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

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

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

I guessss...

wow! thats really clever!!

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

both work like charm! xD

wowwowowow

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