Question

A function has f(x) has a limit L at x = a if and only if it has both left and right limits there and these one­-sided limits are both equal to L.
That is; limx→a f(x) = L ⇔ limx→a f(x) = limx→a+ f(x) = L .
Use this fact about the relationship between one­sided limits and two­sided limits to answer the following questions

a) If f(x)=|x2| / (x^2+x-6) ,find

i. limx→2+ f(x),
ii. limx→2- f (x),
iii. limx→2 f (x),

b) What can you say about the relationship between the solved functions in (a)?

Answer 1

\(\lim\limits_{x \to 2} \dfrac{|x|}{x^2 + x - 6}\)

\(= 2 \lim\limits_{x \to 2} \dfrac{1}{x^2 + x - 6}\)


problem is that the denom goes to zero

we can investigate the left and right-sidedness using a calculator ... or by saying that we let \(x = 2 + \delta\), so \( x- 2 = \delta \), where \( \delta <<1\) and the limit is now \(\delta \to 0\)

\(= 2 \lim\limits_{\delta \to 0} \dfrac{1}{(2+\delta)^2 + 2+ \delta - 6}\)

\(= 2 \lim\limits_{\delta \to 0} \dfrac{1}{5 \delta }\)....... ignoring \(\delta^2\) terms in the denom

You can see that we still get the singularity but that it is \(- \infty \) where \(\delta < 0\), ie the left-sided limit; and \(+ \infty \) where \(\delta > 0\), ie the right-sided limit

IOW

\(\lim\limits_{x \to 2^-} \dfrac{|x|}{x^2 + x - 6} = - \infty\)

\(\lim\limits_{x \to 2^+} \dfrac{|x|}{x^2 + x - 6} = \infty\)

So there is no limit or derivative at \(x = 2\)

Answer 2

it is \(|x^2|\) up top, right?

Answer 3

does that matter ?

Answer 4

oops my bad it is |x-2| on top

Answer 5

If f(x)=|x-2| / (x^2+x-6) ,find

i. limx→2+ f(x),
ii. limx→2- f (x),
iii. limx→2 f (x),

Answer 6

oooh \(|x-2|\) !!

Answer 7

yes

Answer 8

if \(x>2\) then \(|x-2|=x-2\)
if \(x<2\) then \(|x-2|=2-x\)

Answer 9

so it's indeterminate

good spot satellite

you can factor the bottom:

\[\lim\limits_{x \to 2} \dfrac{|x-2|}{(x-2)(x+3)}\]

Answer 10

so the right handed limit will be for \[\frac{x-2}{x^2+x-6}\] and the left handed lmit will be for \[\frac{2-x}{x^2+x-6}\]

Answer 11

factor and cancel, see when you get for the limit
should be pretty clear that one will be the negative of the other

Answer 12

if \(x<2\) you have

\[\lim\limits_{x \to 2^-} \dfrac{-(x-2)}{(x-2)(x+3)}\]

if \(x>2\) you have

\[\lim\limits_{x \to 2^+} \dfrac{x-2}{(x-2)(x+3)}\]

and cos we are playing with limits and never actually getting to 2, you can do the algebra

\[\lim\limits_{x \to 2^-} \dfrac{-(x-2)}{(x-2)(x+3)} \to -\dfrac{1}{2+3}\]

\[\lim\limits_{x \to 2^+} \dfrac{x-2}{(x-2)(x+3)} \to \dfrac{1}{2+3}\]

Answer 13

can i post another question on here?

Answer 14

opening a new thread, and closing this one, is usually best advice :)

Answer 15

ok

Answer 16

relationship between the solved functions? any comments on that?

Answer 17

don't actually understand the question. seems to me there is only 1 function :(