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Limit of the function

wishmaster

Active member
Oct 11, 2013
211
I know i asked similar questions multiple times,but again i have a problem,seems im not good with roots.....
I have to calculate the following limit:

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What do you get if you try substituting $x=0$ into the expression?
 

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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 \(\displaystyle \frac{0}{0}\), 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?
 

wishmaster

Active member
Oct 11, 2013
211
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?
maybe to move the roots somehow into the denumerator?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
To multiply with \(\displaystyle \sqrt{1+x}-\sqrt{1-x}\) ?
No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.
 

wishmaster

Active member
Oct 11, 2013
211
No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.
How?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle 1-x\) ??
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?
 

wishmaster

Active member
Oct 11, 2013
211
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?
\(\displaystyle \sqrt{1+x}+\sqrt{1-x}\) ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \sqrt{1+x}+\sqrt{1-x}\) ?
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:

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

What do you find?
 

wishmaster

Active member
Oct 11, 2013
211
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:

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

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

wishmaster

Active member
Oct 11, 2013
211
This is what you want to multiply the expression with. Since it is $1$, you aren't changing its value. So you want:

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
So i multiply all together?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

wishmaster

Active member
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.

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

Petrus

Well-known member
Feb 21, 2013
739
I know i asked similar questions multiple times,but again i have a problem,seems im not good with roots.....
I have to calculate the following limit:

\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)
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,
\(\displaystyle |\pi\rangle\)

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