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

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SUMMARY

The limit of the expression $$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$ results in an indeterminate form of $$\frac{0}{0}$$. To resolve this, the appropriate method is to multiply the numerator and denominator by the conjugate of the numerator, $$\sqrt{1+x}+\sqrt{1-x}$$. This manipulation simplifies the expression to $$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$, allowing for the cancellation of x and the calculation of the limit.

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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:

$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$
 
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What do you get if you try substituting $x=0$ into the expression?
 
MarkFL said:
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.
 
wishmaster said:
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?
 
MarkFL said:
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?
 
wishmaster said:
maybe to move the roots somehow into the denumerator?

Correct, and how can we accomplish this?
 
MarkFL said:
Correct, and how can we accomplish this?

To multiply with $$\sqrt{1+x}-\sqrt{1-x}$$ ?
 
wishmaster said:
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.
 
MarkFL said:
No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.
How?
 
  • #10
wishmaster said:
How?

What is the conjugate of the numerator?
 
  • #11
MarkFL said:
What is the conjugate of the numerator?

$$1-x$$ ??
Im stupid it seems...
 
  • #12
wishmaster said:
$$1-x$$ ??
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?
 
  • #13
MarkFL said:
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?

$$\sqrt{1+x}+\sqrt{1-x}$$ ?
 
  • #14
wishmaster said:
$$\sqrt{1+x}+\sqrt{1-x}$$ ?

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?
 
  • #15
MarkFL said:
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.
 
  • #16
wishmaster said:
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:

$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$
 
  • #17
MarkFL said:
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?
 
  • #18
wishmaster said:
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.
 
  • #19
MarkFL said:
Yes. You will find you will be able to get rid of the $x$ in the denominator as well when you reduce.
$$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$ ?
 
  • #20
wishmaster said:
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'hospital's 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,
$$|\pi\rangle$$

- - - Updated - - -

wishmaster said:
$$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$ ?
Divide top and bottom with x and Then calculate the limit! Good job!:)
 

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