Question

Let f be defined as follows, where a CANNOT equal 0

f(x) = (x^3 - a^3)/(x - a) when x does not equal a
f(x) = 3a^2 when x DOES equal a

Which of the following are true about f?

I. Lim(x -> a) f(x) exists
II. f(a) exists
III. f(x) is continuous at x = a

Note: It is possible for one, none, or all to be correct

Somebody please help me??? I'm so confused

Answer 1

when \(x\ne a\),\[f(x)=\frac{(x^{3}-a^{3})}{(x-a)}\]when \(x=a\),\[f(x)=3a^{2}\]

Answer 2

This is what you are given, yes?

Answer 3

Yes that is correct (The equations aren't loading on my end but from what I can read that is correct)

Answer 4

Oh, I'm sorry, here's a screenshot of it: http://prntscr.com/d07gqt

Answer 5

Yep that's correct!

Answer 6

Hang on, I'll be right back, I'm going to log out and log back in, hopefully it will help this glitching out, I'll be back in a second!

Answer 7

Okay, well... I've just started learning calculus;
so PLEASE don't take my words as solid proof, but... this is what I think.

(i) cannot be true as the limit at \(a\) is different from both sides. Regarding one-sided functions:\[\lim_{x\rightarrow a^{+}}\ne\lim_{x\rightarrow a^{-}}\] One side would be approaching from \(+\infty\) and the other side would be approaching from \(-\infty\).

(ii) is true as there is a point a \(x=a\), it's stated in the problem.

(iii) cannot be true unless (i) is true, as continuity means that the limit at \(x=a\) needs to exist (\(I\) \(THINK\)).

Answer 8

That's what I thought too! Since I cannot be true, III cannot be true either but II is possible

Answer 9

Yes, well ... I would get a second opinion on this, because I'm not quite sure if I have a full grasp on the concepts yet.

Also, \(\text{Oakheart says hi.}\) 🐈

Answer 10

LOL! Hello Oakheart my passed love XD I'm glad someone else got the name, and I'll do some more research, but that helped a lot, thanks so much!

Answer 11

Alright :-) no problem! Good luck ♣

(And... spoiler alert (?) but Brambleclaw becomes Clan Leader haha)

Answer 12

(Oh oops, I guess so XD)) Oh nu! ;-;

Answer 13

idk how far you are but yeah !

GOOD LUCK ♣

@Directrix @agent0smith help please lol

Answer 14

For part I, you need to factor it to check the left and right hand limits. Use a difference of two cubes (google it) \[\large f(x)=\frac{x^{3}-a^{3}}{x-a}\]

Answer 15

Wait a minute if that is factored that gives me a^2 + ax + x^2 and then when I substitute a for x I get a^2 + aa + a^2, which would equal to 3a^2

Answer 16

Yes, which means what...?

But slow down and show work. You need to find the value of the left and right limits first\[\large \lim_{x \rightarrow a^-}\frac{(x-a)(x^2 +ax + a^2)}{x-a}\]\[\large \lim_{x \rightarrow a^+}\frac{(x-a)(x^2 +ax + a^2)}{x-a}\]are these two going to be equal?

Answer 17

I'm not 100% sure only because I'm not sure how to calculate that and I can't see the equations (my computer is only showing the code) I'm guessing they are going to be equal. I DO know that if I and II are equal to each other that means that they are continuous, so that would mean III is correct as well.

Answer 18

Well, Looking at the equations I would say it's equal.. I'm looking online now to see how to prove this analytically

Answer 19

All you have to do is cancel off common factors, and then it's clear they're equal.

Answer 20

So just factor off the (x-a) and then be left with (x^2 + ax + a^2)

Answer 21

Yep.

Answer 22

Okay, so In reality all of them would be true, correct?

Answer 23

Because 1 and 2 equal the same thing, so 3 is automatically true.

Answer 24

Also thanks for your help XD

Answer 25

Yes, since the limit at a, is equal to f(a), it's continuous.

Answer 26

Perfect, Thanks again for your help.

Answer 27

Welcome.