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?
maybe to move the roots somehow into the denumerator?
Correct, and how can we accomplish this?
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.
How?
What is the conjugate of the numerator?
\(\displaystyle 1-x\) ??
Hey, don't get discouraged...you are learning...it takes time.
\(\displaystyle \sqrt{1+x}+\sqrt{1-x}\) ?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?
Yes, good!
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?
This is what you want to multiply the expression with. Since it is $1$, you aren't changing its value. So you want:
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.
Hello wishmaster!
here you got a Link that explain it well Pauls Online Notes : Calculus I - L'Hospital's Rule and Indeterminate Forms Divide top and bottom with x and Then calculate the limit! Good job!