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#### wishmaster

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- Oct 11, 2013

- 211

I have to calculate the following limit:

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)

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- Oct 11, 2013

- 211

I have to calculate the following limit:

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)

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- Oct 11, 2013

- 211

I get $0$ of course...so i know i should do some operations,so i get rid of the x in the denumerator.What do you get if you try substituting $x=0$ into the expression?

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You actually get \(\displaystyle \frac{0}{0}\), 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?I get $0$ of course...so i know i should do some operations,so i get rid of the x in the denumerator.

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- #5

- Oct 11, 2013

- 211

maybe to move the roots somehow into the denumerator?You actually get \(\displaystyle \frac{0}{0}\), 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?

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Correct, and how can we accomplish this?maybe to move the roots somehow into the denumerator?

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- Oct 11, 2013

- 211

To multiply with \(\displaystyle \sqrt{1+x}-\sqrt{1-x}\) ?Correct, and how can we accomplish this?

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No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.To multiply with \(\displaystyle \sqrt{1+x}-\sqrt{1-x}\) ?

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- Oct 11, 2013

- 211

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

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What is the conjugate of the numerator?How?

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- Oct 11, 2013

- 211

\(\displaystyle 1-x\) ??What is the conjugate of the numerator?

Im stupid it seems.....

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Hey, don't get discouraged...you are learning...it takes time.\(\displaystyle 1-x\) ??

Im stupid it seems.....

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?

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- #13

- Oct 11, 2013

- 211

\(\displaystyle \sqrt{1+x}+\sqrt{1-x}\) ?Hey, don't get discouraged...you are learning...it takes time.

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?

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Yes, good!\(\displaystyle \sqrt{1+x}+\sqrt{1-x}\) ?

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:

\(\displaystyle 1=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)

What do you find?

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- #15

- Oct 11, 2013

- 211

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

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:

\(\displaystyle 1=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)

What do you find?

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- #16

This is what you want to multiply the expression with. Since it is $1$, you aren't changing its value. So you want:But how turned the denumerator into this term? I had $x$ in it.

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)

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- #17

- Oct 11, 2013

- 211

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

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)

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Yes. You will find you will be able to get rid of the $x$ in the denominator as well when you reduce.So i multiply all together?

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- #19

- Oct 11, 2013

- 211

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

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

- Feb 21, 2013

- 739

Hello wishmaster!

I have to calculate the following limit:

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)

I bet it's mather of time until you learn l'hopitals rule which i would use when I see this limit 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,

\(\displaystyle |\pi\rangle\)

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