finding values of 'h' that will make function continuous or not continuous

[avanm]

Hi,

Im having trouble with this. The question states "notice h is not defined at x=0, find the value of h(0), which will make h continuous at x=0, or explain why this is not possible"
I know with determining continuity, the function must following the following conditions
1. The number 'a' is in the domain (ie. does f(a) exists)
2. does the limit_x->af(x) exists
3.limit=f(a)

What I have tried is, I simplified/manipulated the equation by multiplying by the conjugate, and cancelling out the x² on the top and bottom leaving 2+sqrt(4-x²)
then h(x)=2+sqrt(4-x²) and h(0)=4 and the limit_x->0=4 then the limit=f(a) is true, and all the conditions are satisfied.

is my logic correct?

Thanks you

 

[Steven G]

You have the correct answer but I feel that you were very sloppy at how you went about it.

In the h(x) that you were given, h(0) was undefined. The suddenly h(0) =4. That can't be!!! How can h(0) equal both 0 and 4??!!!!

Your new h(x) = the original h(x) at all points except at x=0.

What you needed to write was [MATH]\lim_{x->0}h(x) =\lim_{x->0}(2+\sqrt{4-x^2}) =4 [/MATH]

 

[avanm]

You're saying my answer of it is continuous is correct but if h(0) is undefined and the limit is 4 then x=0 is not continuous? because all the conditions have to be met
also why are you evaluating the limit from the left hand side?

 

[Steven G]

Why do you think that I was evaluating the limit from the left hand side?

To continuity at x=0 you need to verify three conditions.
1a) [MATH]\lim_{x->0^-}h(x) exists[/MATH]1b) [MATH]\lim_{x->0^+}h(x) exists[/MATH]2) [MATH]\lim_{x->0^-}h(x) =\lim_{x->0^+}h(x) =\lim_{x->0}h(x) [/MATH]3)[MATH]\lim_{x->0}h(x) = h(0)[/MATH]
I would simply always very those conditions above to show continuity. If any one of the above conditions do not hold then you have a discontinuity at x=0.

 

[avanm]

Why do you think that I was evaluating the limit from the left hand side?

To continuity at x=0 you need to verify three conditions.
1a) [MATH]\lim_{x->0^-}h(x) exists[/MATH]1b) [MATH]\lim_{x->0^+}h(x) exists[/MATH]2) [MATH]\lim_{x->0^-}h(x) =\lim_{x->0^+}h(x) =\lim_{x->0}h(x) [/MATH]3)[MATH]\lim_{x->0}h(x) = h(0)[/MATH]
I would simply always very those conditions above to show continuity. If any one of the above conditions do not hold then you have a discontinuity at x=0.

Ooooops! I saw the -> symbol and thought it was x^-. my bad!
Okay right, so the left and right hand limits=4 but h(0)= undefined
therefore it isn't continuous at x=0? as per 3) of your post?

 

[Steven G]

Ooooops! I saw the -> symbol and thought it was x^-. my bad!
Okay right, so the left and right hand limits=4 but h(0)= undefined
therefore it isn't continuous at x=0? as per 3) of your post?

What you wrote is true that h(x) is not continuous at x=0.

However the problem asked if it is possible to redefine the function h(x) at x=0 to make the function continuous.

For example if we made h(x) a piecewise function as

[MATH]h(x) = \dfrac{x^2}{2-\sqrt{4-x^2}}, if x\neq0[/MATH] and 4, if x=0.

Now h(x) is continuous at x = 0.

Is this all clear???

 

[JeffM]

You are getting where you need to go, and that probably means your thinking is good, but you are not expressing yourself at all carefully.

[MATH]0 < |x| \le 2 \text { and } f(x) = \dfrac{x^2}{2 - \sqrt{4 - x^2}} = \dfrac{x^2}{2 - \sqrt{4 - x^2}} * 1 =\\ \dfrac{x^2}{2 - \sqrt{4 - x^2}} * \dfrac{2 + \sqrt{4 - x^2}}{2 + \sqrt{4 - x^2}} = \dfrac{x^2(2 + \sqrt{4 - x^2})}{4 - 4 + x^2} = 2 + \sqrt{4 - x^2}.[/MATH]
Your algebra is fine. And f(x) = f(x) * 1 everywhere where f(x) is defined. But multiplying by 1 does not change the domain. However, the new form does show us how to extend the definition of f(x) so that f(x) can be continuous at x = 0. Consider

[MATH]0 \le |x| \le 2 \text { and } g(x) = 2 + \sqrt{4 - x^2}.[/MATH]
At every value of x in the domain of f(x), g(x) = f(x). Moreover, g(x) can be shown to be continuous in (-2, 2) and g(0) clearly equals 4. Thus, you can redefine f(x) so that g(x) = f(x) for all x in the interval [-2, 2] and thereby make the extended f(x) continuous in (- 2, 2) by

[MATH]x = 0 \implies f(x) = 4;\ 0 < |x| \le 2 \implies f(x) = \dfrac{x^2}{2 - \sqrt{4 - x^2}}.[/MATH]
Alternatively, you could just redefine f(x) in the form of g(x).

 

[avanm]

What you wrote is true that h(x) is not continuous at x=0.

However the problem asked if it is possible to redefine the function h(x) at x=0 to make the function continuous.

For example if we made h(x) a piecewise function as

[MATH]h(x) = \dfrac{x^2}{2-\sqrt{4-x^2}}, if x\neq0[/MATH] and 4, if x=0.

Now h(x) is continuous at x = 0.

Is this all clear???

So the original h(x) function is undefined (as the problem states) and were are trying to redefine it so x=0 to make the function continuous.

My new h(x)=(2+√4−x^2), h(0) of my new h(x)=4, so does the left and right hand limits. Therefore my new h(x) meets all the requirements of continuity and is continuous.

What I am confused about is how you said in your initial reply "In the h(x) that you were given, h(0) was undefined. The suddenly h(0) =4. That can't be!!! How can h(0) equal both 0 and 4??!!!!"

How can I write this so that I can explain my new h(x) complies with continuity rules and state h(0)=lim_x->0⁺=lim_0^-=4 WITHOUT implying h(0) in my original h(x) is undefined.

Thank you for explaining this to me, you are so helpful!

 

[avanm]

You are getting where you need to go, and that probably means your thinking is good, but you are not expressing yourself at all carefully.

[MATH]0 < |x| \le 2 \text { and } f(x) = \dfrac{x^2}{2 - \sqrt{4 - x^2}} = \dfrac{x^2}{2 - \sqrt{4 - x^2}} * 1 =\\ \dfrac{x^2}{2 - \sqrt{4 - x^2}} * \dfrac{2 + \sqrt{4 - x^2}}{2 + \sqrt{4 - x^2}} = \dfrac{x^2(2 + \sqrt{4 - x^2})}{4 - 4 + x^2} = 2 + \sqrt{4 - x^2}.[/MATH]
Your algebra is fine. And f(x) = f(x) * 1 everywhere where f(x) is defined. But multiplying by 1 does not change the domain. However, the new form does show us how to extend the definition of f(x) so that f(x) can be continuous at x = 0. Consider

[MATH]0 \le |x| \le 2 \text { and } g(x) = 2 + \sqrt{4 - x^2}.[/MATH]
At every value of x in the domain of f(x), g(x) = f(x). Moreover, g(x) can be shown to be continuous in (-2, 2) and g(0) clearly equals 4. Thus, you can redefine f(x) so that g(x) = f(x) for all x in the interval [-2, 2] and thereby make the extended f(x) continuous in (- 2, 2) by

[MATH]x = 0 \implies f(x) = 4;\ 0 < |x| \le 2 \implies f(x) = \dfrac{x^2}{2 - \sqrt{4 - x^2}}.[/MATH]
Alternatively, you could just redefine f(x) in the form of g(x).

Would you be able to view my most recent reply to Jomo?
Would I just refer to my new h(x) as g(x), im having trouble labeling the two as separate functions.
Also would I say that h(x)=g(x) do I need to explain why?

thank u!!!

 

[Steven G]

I hope that it is clear that you just can't have two different h(x)'s.

Now if x is not 0, then h(x) =[MATH]2+\sqrt{4-x^2}[/MATH]
[MATH]\lim_{x->0^-}h(x) =\lim_{x->0^-}(2+\sqrt{4-x^2}) =4 [/MATH] and [MATH]\lim_{x->0^+}h(x) =\lim_{x->0+}(2+\sqrt{4-x^2}) =4[/math]
So we showed that the two limits exists (since they equal real numbers) and that they are equal to one another (they both equaled 4).

The last step is to show that h(0) = 4. It does because we defined h(0) =4!!

Note that I never introduced new functions or at least I never named them.

 

[Steven G]

Would you be able to view my most recent reply to Jomo?
Would I just refer to my new h(x) as g(x), im having trouble labeling the two as separate functions.
Also would I say that h(x)=g(x) do I need to explain why?

thank u!!!

Do not define a new function! If you do, then do NOT say that h(x) = g(x) unless they are equal EVERYWHERE!

 

[Steven G]

You have the correct answer but I feel that you were very sloppy at how you went about it.

In the h(x) that you were given, h(0) was undefined. The suddenly h(0) =4. That can't be!!! How can h(0) equal both 0 and 4??!!!!

Your new h(x) = the original h(x) at all points except at x=0.

What you needed to write was [MATH]\lim_{x->0}h(x) =\lim_{x->0}(2+\sqrt{4-x^2}) =4 [/MATH]

Meant to say How can h(0) equal both undefined and 4??!!!!

 

[avanm]

I hope that it is clear that you just can't have two different h(x)'s.

Now if x is not 0, then h(x) =[MATH]2+\sqrt{4-x^2}[/MATH]
[MATH]\lim_{x->0^-}h(x) =\lim_{x->0^-}(2+\sqrt{4-x^2}) =4 [/MATH] and [MATH]\lim_{x->0^+}h(x) =\lim_{x->0+}(2+\sqrt{4-x^2}) =4[/math]
So we showed that the two limits exists (since they equal real numbers) and that they are equal to one another (they both equaled 4).

The last step is to show that h(0) = 4. It does because we defined h(0) =4!!

Note that I never introduced new functions or at least I never named them.

Okay this makes sense now, im over complicating it in my brain haha! thank you so much you are incredibly helpful!