Limit, unknown

[Miriams]

Find the number a so that the limit
exists. Also find the limit for this situation.

Anyone who can explain how to find the answer?

 

[MarkFL]

Hello, and welcome to FMH!

In order for the limit to exist, we need for the numerator to approach zero as \(x\to7\). Can you find what \(a\) must be in order for that to happen?

 

[pka]

Find the number a so that the limit
View attachment 13910exists. Also find the limit for this situation.

Anyone who can explain how to find the answer?

Miriams, is the given limit \(\displaystyle \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + a} - 2\sqrt x }}{{1 - x}}\)
OR IS it \(\displaystyle \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {{x^2} + a} - 2\sqrt x }}{{7 - x}}\)

 

[pka]

Which ever case it is, multiply numerator & denominator by \(\displaystyle \sqrt{x^2+a}+2\sqrt{3}\).
Then choose \(\displaystyle a=?\) so that \(\displaystyle (x^2+a)-4x\) has a factor of \(\displaystyle (x-1) or (x-7)\)

 

[HallsofIvy]

The first thing I would do is note that, if we simply substitute x= 1, the denominator is 0. In order that this limit exist the numerator must also be 0. We must have \(\displaystyle \sqrt{1+ a}- 2= 0\). So \(\displaystyle \sqrt{1+ a}= 2\), 1+ a= 4, a= 3. Now do as pka suggested- multiply both numerator and denominator by \(\displaystyle \sqrt{x^2+ 3}+ 2\sqrt{x}\). (His final \(\displaystyle \sqrt{3}\)" appears to be a typo.)

 

[Miriams]

Miriams, is the given limit \(\displaystyle \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + a} - 2\sqrt x }}{{1 - x}}\)
OR IS it \(\displaystyle \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {{x^2} + a} - 2\sqrt x }}{{7 - x}}\)

It is the one where x->1

 

[pka]

Thank you for the clarification. Now
\(\displaystyle \frac{\sqrt{x^2+a}-2\sqrt x}{1-x}\left(\frac{\sqrt{x^2+a}+2\sqrt x}{\sqrt{x^2+a}+2\sqrt x}\right)=\frac{(x^2+a)-4x}{(1-x)(\sqrt{x^2+a}+2\sqrt x)}\)
Now choose a value for \(\displaystyle a\) so you can factor \(\displaystyle x^2-4x+a=(x-1)(~?~)\)
Divide off the \(\displaystyle (x-1)\) and find the limit.

 

[Miriams]

Thank you for the clarification. Now
\(\displaystyle \frac{\sqrt{x^2+a}-2\sqrt x}{1-x}\left(\frac{\sqrt{x^2+a}+2\sqrt x}{\sqrt{x^2+a}+2\sqrt x}\right)=\frac{(x^2+a)-4x}{(1-x)(\sqrt{x^2+a}+2\sqrt x)}\)
Now choose a value for \(\displaystyle a\) so you can factor \(\displaystyle x^2-4x+a=(x-1)(~?~)\)
Divide off the \(\displaystyle (x-1)\) and find the limit.

Hmm, I am not sure what you meant in the last step. So we do not use L'hopitals rule?

 

[pka]

Hmm, I am not sure what you meant in the last step.
So we do not use L'hopitals rule? Absolutely not!

The limit is \(\displaystyle x\to 1\) so we have \(\displaystyle 0\) in the denominator. Thus we need to rid yourself of it.
So what value of \(\displaystyle a\) would give us a factor \(\displaystyle (x-1)\) in the numerator?
If you have \(\displaystyle (x-1)\) in the numerator and \(\displaystyle (1-x)\) in the denominator thy divide to \(\displaystyle -1\).
Now as \(\displaystyle x\to 1\) the limit is defined. So \(\displaystyle a=~?\).
Do some work for yourself. We are not here to do your work for you.

 

[MarkFL]

It is the one where x->1

Okay, so we need:

[MATH]\sqrt{1^2+a}-2\sqrt{1}=0[/MATH]
[MATH]\sqrt{a+1}=2[/MATH]
[MATH]a+1=4[/MATH]
[MATH]a=3[/MATH]
So we now have:

[MATH]L=\lim_{x\to1}\frac{\sqrt{x^2+3}-2\sqrt{x}}{1-x}[/MATH]
We could use L'Hôpital's rule since we have the indeterminate form 0/0, but let's rationalize the numerator:

[MATH]L=\lim_{x\to1}\frac{\sqrt{x^2+3}-2\sqrt{x}}{1-x}\cdot\frac{\sqrt{x^2+3}+2\sqrt{x}}{\sqrt{x^2+3}+2\sqrt{x}}[/MATH]
[MATH]L=\lim_{x\to1}\frac{x^2-4x+3}{(1-x)(\sqrt{x^2+3}+2\sqrt{x})}[/MATH]
[MATH]L=\lim_{x\to1}\frac{(x-1)(3-x)}{(x-1)(\sqrt{x^2+3}+2\sqrt{x})}[/MATH]
[MATH]L=\lim_{x\to1}\frac{3-x}{\sqrt{x^2+3}+2\sqrt{x}}[/MATH]
Can you proceed?

 

[Miriams]

Okay, so we need:

[MATH]\sqrt{1^2+a}-2\sqrt{1}=0[/MATH]
[MATH]\sqrt{a+1}=2[/MATH]
[MATH]a+1=4[/MATH]
[MATH]a=3[/MATH]
So we now have:

[MATH]L=\lim_{x\to1}\frac{\sqrt{x^2+3}-2\sqrt{x}}{1-x}[/MATH]
We could use L'Hôpital's rule since we have the indeterminate form 0/0, but let's rationalize the numerator:

[MATH]L=\lim_{x\to1}\frac{\sqrt{x^2+3}-2\sqrt{x}}{1-x}\cdot\frac{\sqrt{x^2+3}+2\sqrt{x}}{\sqrt{x^2+3}+2\sqrt{x}}[/MATH]
[MATH]L=\lim_{x\to1}\frac{x^2-4x+3}{(1-x)(\sqrt{x^2+3}+2\sqrt{x})}[/MATH]
[MATH]L=\lim_{x\to1}\frac{(x-1)(3-x)}{(x-1)(\sqrt{x^2+3}+2\sqrt{x})}[/MATH]
[MATH]L=\lim_{x\to1}\frac{3-x}{\sqrt{x^2+3}+2\sqrt{x}}[/MATH]
Can you proceed?

I tried this, but if I substitute the x with 1, there is no longer zero in the numerator and the denominator

 

[MarkFL]

I don't know where the expression you're working with came from.

If you continue from where I left off, you will find:

[MATH]L=\frac{3-1}{\sqrt{1^2+3}+2\sqrt{1}}=\frac{2}{2+2}=\frac{1}{2}[/MATH]

 

[tkhunny]

"unknown" - This is a troublesome word. It would have to mean that there IS a limit, but you just don't know it.

This is very much different from "there is no limit" or "the limit does not exist".

 

[Miriams]

"unknown" - This is a troublesome word. It would have to mean that there IS a limit, but you just don't know it.

This is very much different from "there is no limit" or "the limit does not exist".

Yes I agree with you. I meant the number is unknown. I don't know how to change it.

 

[Miriams]

I don't know where the expression you're working with came from.

If you continue from where I left off, you will find:

[MATH]L=\frac{3-1}{\sqrt{1^2+3}+2\sqrt{1}}=\frac{2}{2+2}=\frac{1}{2}[/MATH]

Yes, that is the same as I got when I exchanged the x with 1. But is this the final answer? We didn't need to derivate?

 

[MarkFL]

Yes, that is the same as I got when I exchanged the x with 1. But is this the final answer? We didn't need to derivate?

Yes, that's the value of the limit. Like I said, we could also have applied L'Hôpital's rule:

[MATH]L=\lim_{x\to1}\frac{\sqrt{x^2+3}-2\sqrt{x}}{1-x}=\lim_{x\to1}\frac{\dfrac{x}{\sqrt{x^2+3}}-\dfrac{1}{\sqrt{x}}}{-1}=\frac{1}{2}[/MATH]
But, since we algebraically manipulated the expression so that it was no longer an indeterminate form, we could use direct substitution.

 

[Miriams]

[MATH][/MATH]

Yes, that's the value of the limit. Like I said, we could also have applied L'Hôpital's rule: [MATH]L=\lim_{x\to1}\frac{\sqrt{x^2+3}-2\sqrt{x}}{1-x}=\lim_{x\to1}\frac{\dfrac{x}{\sqrt{x^2+3}}-\dfrac{1}{\sqrt{x}}}{-1}=\frac{1}{2}[/MATH] But, since we algebraically manipulated the expression so that it was no longer an indeterminate form, we could use direct substitution. Oh, now I understand. Thank you so much for the help!

 

[HallsofIvy]

I tried this, but if I substitute the x with 1, there is no longer zero in the numerator and the denominator

?? That's a GOOD thing, isn't it? You don't want a zero in the denominator!

 

[tkhunny]

Why are you substituting anything? Did you prove it was continuous?