Limit of function containing absolute value

gimme5minutes

OK I have another one of these questions. I think I have the correct answer but Im not sure how I am supposed to give the final answer -

I have to prove that
[latex] \lim_{x \to 2} \ \frac{|x \ - \ 2|}{x \ - \ 2}[/latex]

does not exist.

Well here's what I've done -
Take left hand side -
[latex] \lim_{x \to 2^-} \ \frac{|x \ - \ 2|}{x \ - \ 2}[/latex]

[latex] \lim_{x \to 2^-} \ \frac{-(x \ - \ 2)}{x \ - \ 2}[/latex]


[latex] \lim_{x \to 2^-} \ \frac{2 \ - \ x}{x \ - \ 2}[/latex]

Take right hand side -
[latex] \lim_{x \to 2^+} \ \frac{|x \ - \ 2|}{x \ - \ 2}[/latex]

[latex] \lim_{x \to 2^+} \ \frac{x \ - \ 2}{x \ - \ 2}[/latex]


I'm not sure what to do with the answers for the left and right hand sides. I can see that from the left hand side as x approaches 2 there will be a positive numerator and a negative denominator. Whereas when x approaches 2 from the right hand side there will be a positive numerator and denominator. There the limit on the right hand side will be positive whereas the left hand side will be negative, therefore they are not equal...But what is the correct way of writing that in maths terms?

gimme5minutes

Taking the right hand side....can I say that

[latex] \lim_{x \to 2^+} \ \frac{x \ - \ 2}{x \ - \ 2} = 1[/latex]

as x is close to 2 but not actually equal to 2.

And similary with the left hand side, it is equal to -1. And [latex]-1 $\neq$ 1[/latex], therefore the limit does not exist. Is that correct?

Fremen

In both cases, if you substitute x=2 in, you end up with 0/0. This means you can apply L'hopital's rule. It will tell you that the limit on one side is 1, while the limit on the other side is -1.

MathsManiac

gimme5minutes wrote: »

Taking the right hand side....can I say that

[latex] \lim_{x \to 2^+} \ \frac{x \ - \ 2}{x \ - \ 2} = 1[/latex]

as x is close to 2 but not actually equal to 2.

And similary with the left hand side, it is equal to -1. And [latex]-1 $\neq$ 1[/latex], therefore the limit does not exist. Is that correct?

Yes, you can say that, (so you don't need to use l'Hôpital's rule, even though you can).

Since, for all x > 2, it is true to say that (x-2)/(x-2)=1, it follows that the limit as x approaches 2 from above is the same as the limit of the function f(x)=1 as x approaches 2 from above, which is 1.