Evaluation of Limit of Piecewise function help needed

[kashan123999]

Homework Statement

Find the continuity of a piece wise function f(x) = (x^2-9)/(x-3) for x not equal to 3 and f(x) = 6 for x=3...the continuity is to be found at x = 3

Homework Equations

Limit f(x) as x--> c = L

The Attempt at a Solution

By seeing the first rule it is clearly written on the wall that the function is discontinuous but the answer on my course book is continuous...can anyone explain that please...I know ok f(3) = 6 so it is defined at f of 3 but for the first rule it is not defined,how we are supposed to solve it? please give me a Thorough explanation

 

[Ray Vickson]

Homework Statement

Find the continuity of a piece wise function f(x) = x^2-9/x-3 for x not equal to 3 and f(x) = 6 for x=3...the continuity is to be found at x = 3

Homework Equations

Limit f(x) as x--> c = L

The Attempt at a Solution

By seeing the first rule it is clearly written on the wall that the function is discontinuous but the answer on my course book is continuous...can anyone explain that please...I know ok f(3) = 6 so it is defined at f of 3 but for the first rule it is not defined,how we are supposed to solve it? please give me a Thorough explanation

As in your previous post, you have written
$$f(x) = x^2 -\frac{9}{x} -3 \; \text{ if } x \neq 3$$
and $f(3) = 6$.

To test for continuity, just evaluate $\lim_{x \to 3} f(x)$ and compare it to f(3).

The question would be very different if you had written (x^2-9)/(x-3), which is
$$\frac{x^2-9}{x-3}$$
but since you did not write this I assume you did not mean this.

 

[kashan123999]

how is it now :D:D

 

[Ray Vickson]

how is it now :D:D

It is good now.

You said "By seeing the first rule it ... ", but we don't know what the first rule says.

What you need to do is check two things: (i) does $\lim_{x \to 3} f(x)$ exist?; and (ii) if the limit exists, is it equal to f(3)?

So, can you do step (i)? Of course, you cannot just put x = 3 in the formula for f(x), since that will give you the illegal/indeterminate/meaningless form 0/0, but in calculus one studies various ways to deal with such situations. Have you taken that material yet?

If you have not yet studied the material I referred to, can you at least re-write the formula for f(x) so that $f(x) = f_1(x)$ for $x \neq 3$, and $\lim_{x \to 3} f_1(x)$ is easier to get? Can you see why $\lim f(x) = \lim f_1(x)$?

 

[sassafrasaxe]

When stuff like this happens where you evaluate (x^2 - 9)/(x-3) to be 0/0, you can try to factor out the expressions.
Here, the numerator factors out pretty easily. (x^2 - 9) becomes (x+3)(x-3), so then you have
[(x+3)(x-3) / (x-3)]
Now just cancel out the 2 (x-3)'s, giving you (x+3)/1.
Now substitute that 3 in for x, since you want to evaluate it at x = 3, or more specifically, as x approaches 3.
That gives you (3+3)/1 = 6.
Since it's asking you if the value that it gives for x = 3 (being 6) is continuous, to find out if it is, you must compare the value that it gives you with the value that the function (x^2 - 9)/(x-3) actually approaches.
By factoring out and canceling we already found that the value approaches 6, which is what they assigned the value at that x to be.
Since they didn't assign it to another value (i.e. 12, in which case it would be getting closer and closer to 6, then when it got to 6 it would go to 12, then it would go farther and farther from 6, starting infinitely close, this being discontinuous),
but since it's NOT 12, and IS indeed 6, it is continuous, because there are no jump discontinuities such as that.

A quicker approach that you'll learn later is "L'Hopital's rule", in which you take the derivative of the numerator and the denominator if it is indeterminate form, until you have something that is not indeterminate form. (0/0)

 

[kashan123999]

It is good now.

You said "By seeing the first rule it ... ", but we don't know what the first rule says.

What you need to do is check two things: (i) does $\lim_{x \to 3} f(x)$ exist?; and (ii) if the limit exists, is it equal to f(3)?

So, can you do step (i)? Of course, you cannot just put x = 3 in the formula for f(x), since that will give you the illegal/indeterminate/meaningless form 0/0, but in calculus one studies various ways to deal with such situations. Have you taken that material yet?

If you have not yet studied the material I referred to, can you at least re-write the formula for f(x) so that $f(x) = f_1(x)$ for $x \neq 3$, and $\lim_{x \to 3} f_1(x)$ is easier to get? Can you see why $\lim f(x) = \lim f_1(x)$?

Ok i know we can factorize and all that stuff and then just plug in x = 3 and the answer will be 6 which clearly shows that the function is continous...in another problem which i encountered...there was 1 rule only stating that f(x) = (x^2-1)/(x-1) discuss continuity at x = 1... i KNOW apparently it is not defined but by factorizing... x+1 will remain...plugging x will give us 2...but on the answer's manual the function was defined as discontinous? wtf is this i mean...Couldn't get the idea at all...

similarly removing the 2nd rule from the thread's question...we get f(x) = (x^2-9)/(x-3) and it is discontinous at x= 3 :( why is that so,we can factorize it and THEN put the x = 3 and say the limit is again 6...why is that we won't factorize it??

 

[Ray Vickson]

Ok i know we can factorize and all that stuff and then just plug in x = 3 and the answer will be 6 which clearly shows that the function is continous...in another problem which i encountered...there was 1 rule only stating that f(x) = (x^2-1)/(x-1) discuss continuity at x = 1... i KNOW apparently it is not defined but by factorizing... x+1 will remain...plugging x will give us 2...but on the answer's manual the function was defined as discontinous? wtf is this i mean...Couldn't get the idea at all...

similarly removing the 2nd rule from the thread's question...we get f(x) = (x^2-9)/(x-3) and it is discontinous at x= 3 :( why is that so,we can factorize it and THEN put the x = 3 and say the limit is again 6...why is that we won't factorize it??

Please stop talking about Rule 1, etc., which are written on the wall. We cannot see the wall, and have no idea what you are talking about. Just tell us in words what you are trying to say.

 

[HallsofIvy]

I will say what Ray Vickson said in his first response to your post: IF this function were defined just by $f(x)= (x^2- 9)/(x- 3)$ THEN it would not be continuous at x= 3 because it is not DEFINED there.
A function, f(x), is defined to be "continuous at x= a" if and only if three things are true:
1) f(a) is defined
2) [itex]\lim_{x\to a} f(x) exist
3) $\lim_{x\to a} f(x)= f(a)$

Or course, for that third to make any sense, the first two must be true so most often we only mention the third rule.
Here,, f(3) is defined by f(3)= 6 so the question of "continuity" is just "Does the limit, as x goes to 3, of f(x) exist and is it equal to 6?" Can you answer that question?

 

[kashan123999]

I will say what Ray Vickson said in his first response to your post: IF this function were defined just by $f(x)= (x^2- 9)/(x- 3)$ THEN it would not be continuous at x= 3 because it is not DEFINED there.
A function, f(x), is defined to be "continuous at x= a" if and only if three things are true:
1) f(a) is defined
2) [itex]\lim_{x\to a} f(x) exist
3) $\lim_{x\to a} f(x)= f(a)$

Or course, for that third to make any sense, the first two must be true so most often we only mention the third rule.
Here,, f(3) is defined by f(3)= 6 so the question of "continuity" is just "Does the limit, as x goes to 3, of f(x) exist and is it equal to 6?" Can you answer that question?

yes it seems to exist :) now i understand cause the 2nd rules states that so...if we put x=3 so f(3) = 6 hence x exists at 3 where it was previously apparently seemed undefined by the previous rule...hence the limit exists...thank you sir :)