Finding values of 'a' that makes the function continuous

[avanm]

Hello!

I'm trying to figure out the values of a make f continuous everywhere.
Normally what I would do if there was an 'a', I would find the left and right hand limits at the 'a' value then I would be left with a equation to solve to get the 'a' value.
So here I attempted to solve for the limit x->a for both equations and I got lim_x->a^-= 1/2sqrt(a) and im_x->a⁺= 1/sqrt(3a²+1)

now I dont know what to do, your help is very much appreciated!

 

[Steven G]

Set the two limits equal to one another and solve for a.

 

[JeffM]

[MATH]x < a \text { and } \dfrac{\sqrt{x} - \sqrt{a}}{x - a} = \dfrac{\cancel {(\sqrt{x} - \sqrt{a})}}{\cancel {(\sqrt{x} - \sqrt{a})}(\sqrt{x} + \sqrt{a})} = \dfrac{1}{\sqrt{x} + \sqrt{a}}.[/MATH]
[MATH]x = a \implies \dfrac{1}{\sqrt{x} + \sqrt{a}} = \dfrac{1}{2\sqrt{a}} \text { and} \dfrac{1}{\sqrt{3x^2 + 1}} = \dfrac{1}{\sqrt{3a^2 + 1}}.[/MATH]
[MATH]\dfrac{1}{2\sqrt{a}} = \dfrac{1}{\sqrt{3a^2 + 1}} \iff \dfrac{1}{4a} = \dfrac{1}{3a^2 + 1} \iff \\ 3a^2 - 4a + 1 = 0 \iff a = \dfrac{4 \pm \sqrt{16 - 4 * 3 * 1}}{6} = \dfrac{1}{3} \text { or } 1.[/MATH]Now to prove this, you have to do limits. But now you know what you are aiming for.

I find it easier to fuss about limits until after I've done the algebra.

 

[Steven G]

[MATH]x < a \text { and } \dfrac{\sqrt{x} - \sqrt{a}}{x - a} = \dfrac{\cancel {(\sqrt{x} - \sqrt{a})}}{\cancel {(\sqrt{x} - \sqrt{a})}(\sqrt{x} + \sqrt{a})} = \dfrac{1}{\sqrt{x} + \sqrt{a}}.[/MATH]
[MATH]x = a \implies \dfrac{1}{\sqrt{x} + \sqrt{a}} = \dfrac{1}{2\sqrt{a}} \text { and} \dfrac{1}{\sqrt{3x^2 + 1}} = \dfrac{1}{\sqrt{3a^2 + 1}}.[/MATH]
[MATH]\dfrac{1}{2\sqrt{a}} = \dfrac{1}{\sqrt{3a^2 + 1}} \iff \dfrac{1}{4a} = \dfrac{1}{3a^2 + 1} \iff \\ 3a^2 - 4a + 1 = 0 \iff a = \dfrac{4 \pm \sqrt{16 - 4 * 3 * 1}}{6} = \dfrac{1}{3} \text { or } 1.[/MATH]Now to prove this, you have to do limits. But now you know what you are aiming for.

I find it easier to fuss about limits until after I've done the algebra.

Jeff, why is finding those values for a not enough? I'm a bit tired right now but I do not see where things can go wrong (in general).

 

[avanm]

Jeff, why is finding those values for a not enough? I'm a bit tired right now but I do not see where things can go wrong (in general).

[MATH]x < a \text { and } \dfrac{\sqrt{x} - \sqrt{a}}{x - a} = \dfrac{\cancel {(\sqrt{x} - \sqrt{a})}}{\cancel {(\sqrt{x} - \sqrt{a})}(\sqrt{x} + \sqrt{a})} = \dfrac{1}{\sqrt{x} + \sqrt{a}}.[/MATH]
[MATH]x = a \implies \dfrac{1}{\sqrt{x} + \sqrt{a}} = \dfrac{1}{2\sqrt{a}} \text { and} \dfrac{1}{\sqrt{3x^2 + 1}} = \dfrac{1}{\sqrt{3a^2 + 1}}.[/MATH]
[MATH]\dfrac{1}{2\sqrt{a}} = \dfrac{1}{\sqrt{3a^2 + 1}} \iff \dfrac{1}{4a} = \dfrac{1}{3a^2 + 1} \iff \\ 3a^2 - 4a + 1 = 0 \iff a = \dfrac{4 \pm \sqrt{16 - 4 * 3 * 1}}{6} = \dfrac{1}{3} \text { or } 1.[/MATH]Now to prove this, you have to do limits. But now you know what you are aiming for.

I find it easier to fuss about limits until after I've done the algebra.

So I got the same x=1 and =1/3 values, do I have to solve for the limits to see which one goes where? like if 1 is for x<1 or x>=1?

 

[Steven G]

So I got the same x=1 and =1/3 values, do I have to solve for the limits to see which one goes where? like if 1 is for x<1 or x>=1?

I do not see why you would have to check but JeffM clearly stated that you should. I respect JeffM greatly so I am assuming that maybe I am missing something. JeffM or someone else will clear this up with a post.

I must say that I do not think that you should be saying that x=1 or 1/3 but rather that a = 1 or 1/3.

I see you question now. No, no! For both parts of the equation you use a=1 or for both parts of the equation or you use a=1/3. It really is just one equation and you can't substitute in different values for a! Besides if you did then the function would NOT be continuous. Check for yourself and see!

 

[JeffM]

Jeff, why is finding those values for a not enough? I'm a bit tired right now but I do not see where things can go wrong (in general).

Nothing can go wrong, but I ignored all the technical requirements for showing continuity. I am not sure how formal the OP must be. I did not do the stuff about proving existence when x = a and proving the right and left limits equal the value of the function at a. I agree it is common sense, but the OP may have a fair amount of formalism yet to do. Sorry for causing confusion. I have been working on boring stuff for hours.

 

[Steven G]

Replace all a's with 1 and see if the function is continuous at x=1.

Then replace all a's with 1/3 and see if the function is continuous at x=1/3

 

[Steven G]

Nothing can go wrong, but Iignored all the technical requirements for showing continuity. I am not sure how formal the OP must be. I did not do the stuff about proving existence when x = a and proving the right and left limits equal the value of the function at a. I agree it is common sense, but the OP may have a fair amount of formalism yet to do.

Thanks.
I don't see your point about ignoring all technical requirements for showing continuity.
The OP found the left hand limit and found the right hand limit. Then we instructed the OP to set these limits equal to one another. We found that a=1 and a=1/3 made the left hand and right hand limits equal to one another. Ah, the OP has to now show the last part for continuity! That is that f(a) = limit. Good catch. I missed that one!

To OP: So far we found values for a to make the left hand limit = right hand limit. That only shows that the limit exists. To get continuity you now have to show that f(a) = that common limit. Of course it does but you have to state it! I knew JeffM had a reason for saying what he said. Once again I am guilty of sloppy thinking.

 

[avanm]

Thanks.
I don't see your point about ignoring all technical requirements for showing continuity.
The OP found the left hand limit and found the right hand limit. Then we instructed the OP to set these limits equal to one another. We found that a=1 and a=1/3 made the left hand and right hand limits equal to one another. Ah, the OP has to now show the last part for continuity! That is that f(a) = limit. Good catch. I missed that one!

To OP: So far we found values for a to make the left hand limit = right hand limit. That only shows that the limit exists. To get continuity you now have to show that f(a) = that common limit. Of course it does but you have to state it! I knew JeffM had a reason for saying what he said. Once again I am guilty of sloppy thinking.

Nothing can go wrong, but I ignored all the technical requirements for showing continuity. I am not sure how formal the OP must be. I did not do the stuff about proving existence when x = a and proving the right and left limits equal the value of the function at a. I agree it is common sense, but the OP may have a fair amount of formalism yet to do. Sorry for causing confusion. I have been working on boring stuff for hours.

Awesome thank you, I found that a=1 and 1/3 are both continuous !

 

[Steven G]

Awesome thank you, I found that a=1 and 1/3 are both continuous !

Not quite correct. a= 1 and a= 1/3--they are just numbers. Numbers can't be continuous! Functions can. What you meant to say is that f(x) is continuous at x=a if a=1 or a=1/3.

Who said that math is not a language!