Question

How do I start with questions that look like this?:

Answer 1

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

Answer 2

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.

Answer 3

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).

Answer 4

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

Answer 5

you're right. I missed that

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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)

Answer 14

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

Answer 15

what this ?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

I guessss...

Answer 20

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

Answer 21

wow! thats really clever!!

Answer 22

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

Answer 23

both work like charm! xD

Answer 24

wowwowowow

Answer 25

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

Answer 26

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"

Answer 27

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