MHB What is the limit of (sqrt(1+x)-sqrt(1-x))/x as x approaches 0?

[theakdad]

I know i asked similar questions multiple times,but again i have a problem,seems I am not good with roots...
I have to calculate the following limit:

$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$

 

[MarkFL]

What do you get if you try substituting $x=0$ into the expression?

 

[theakdad]

What do you get if you try substituting $x=0$ into the expression?

I get $0$ of course...so i know i should do some operations,so i get rid of the x in the denumerator.

 

[MarkFL]

I get $0$ of course...so i know i should do some operations,so i get rid of the x in the denumerator.

You actually get $$\frac{0}{0}$$, and this is a dreaded indeterminate form. So, since you have seen problems like this before, what would you say we need to do to get it into a determinate form? What would be a good strategy?

 

[theakdad]

You actually get $$\frac{0}{0}$$, and this is a dreaded indeterminate form. So, since you have seen problems like this before, what would you say we need to do to get it into a determinate form? What would be a good strategy?

maybe to move the roots somehow into the denumerator?

 

[MarkFL]

maybe to move the roots somehow into the denumerator?

Correct, and how can we accomplish this?

 

[theakdad]

Correct, and how can we accomplish this?

To multiply with $$\sqrt{1+x}-\sqrt{1-x}$$ ?

 

[MarkFL]

To multiply with $$\sqrt{1+x}-\sqrt{1-x}$$ ?

No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.

 

[theakdad]

No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.

How?

 

[MarkFL]

How?

What is the conjugate of the numerator?

 

[theakdad]

What is the conjugate of the numerator?

$$1-x$$ ??
Im stupid it seems...

 

[MarkFL]

$$1-x$$ ??
Im stupid it seems...

Hey, don't get discouraged...you are learning...it takes time. :D

Consider the expression $a+b$. It's conjugate is $a-b$. Do you see that if we multiply them together, we will have a difference of squares, and a squared radical is no loger a radical. So now what would you say the conjugate of the numerator is?

 

[theakdad]

Hey, don't get discouraged...you are learning...it takes time. :D

Consider the expression $a+b$. It's conjugate is $a-b$. Do you see that if we multiply them together, we will have a difference of squares, and a squared radical is no loger a radical. So now what would you say the conjugate of the numerator is?

$$\sqrt{1+x}+\sqrt{1-x}$$ ?

 

[MarkFL]

$$\sqrt{1+x}+\sqrt{1-x}$$ ?

Yes, good! :D

Now, you want to multiply the expression for which you are asked to find the limit by $1$ in the form of this conjugate divided by itself:

$$1=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$

What do you find?

 

[theakdad]

Yes, good! :D

Now, you want to multiply the expression for which you are asked to find the limit by $1$ in the form of this conjugate divided by itself:

$$1=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$

What do you find?

But how turned the denumerator into this term? I had $x$ in it.

 

[MarkFL]

But how turned the denumerator into this term? I had $x$ in it.

This is what you want to multiply the expression with. Since it is $1$, you aren't changing its value. So you want:

$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$

 

[theakdad]

This is what you want to multiply the expression with. Since it is $1$, you aren't changing its value. So you want:

$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$

So i multiply all together?

 

[MarkFL]

So i multiply all together?

Yes. You will find you will be able to get rid of the $x$ in the denominator as well when you reduce.

 

[theakdad]

Yes. You will find you will be able to get rid of the $x$ in the denominator as well when you reduce.

$$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$ ?

 

[Petrus]

I know i asked similar questions multiple times,but again i have a problem,seems I am not good with roots...
I have to calculate the following limit:

$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$

Hello wishmaster!
I bet it's mather of time until you learn l'hopitals rule which i would use when I see this limit :P here you got a Link that explain it well Pauls Online Notes : Calculus I - L'Hospital's Rule and Indeterminate Forms
Ofc MHB Will help you if you need help understanding it:)
Have a nice weekend!:)

Regards,
$$|\pi\rangle$$

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$$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$ ?

Divide top and bottom with x and Then calculate the limit! Good job!:)