MarkFL said:What do you get if you try substituting $x=0$ into the expression?
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.
MarkFL said: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 said:maybe to move the roots somehow into the denumerator?
MarkFL said:Correct, and how can we accomplish this?
wishmaster said:To multiply with \(\displaystyle \sqrt{1+x}-\sqrt{1-x}\) ?
How?MarkFL said:No, you want to use the conjugate of the numerator, this way the radicals will disappear from the numerator.
wishmaster said:How?
MarkFL said:What is the conjugate of the numerator?
wishmaster said:\(\displaystyle 1-x\) ??
Im stupid it seems...
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?
wishmaster said:\(\displaystyle \sqrt{1+x}+\sqrt{1-x}\) ?
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:
\(\displaystyle 1=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
What do you find?
wishmaster said:But how turned the denumerator into this term? I had $x$ in it.
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:
\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
wishmaster said:So i multiply all together?
\(\displaystyle \frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}\) ?MarkFL said:Yes. You will find you will be able to get rid of the $x$ in the denominator as well when you reduce.
Hello wishmaster!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:
\(\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)
Divide top and bottom with x and Then calculate the limit! Good job!:)wishmaster said:\(\displaystyle \frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}\) ?
A limit of a function is a mathematical concept that describes the behavior of a function as its input approaches a certain value. It is the value that the function gets closer and closer to as the input gets closer and closer to the specified value.
The limit of a function can be determined by evaluating the function at values that are closer and closer to the specified value. If the function values approach a specific number as the input values get closer to the specified value, then that number is the limit of the function.
Limits are important in calculus because they allow us to analyze the behavior of a function at a specific point without having to actually evaluate the function at that point. This allows us to find the behavior of a function at points where it may not be defined or where it may be difficult to evaluate.
A left-hand limit is the value that a function approaches as the input values approach from the left side of the specified value. Similarly, a right-hand limit is the value that a function approaches as the input values approach from the right side of the specified value. The main difference is the direction from which the function values are approaching the specified value.
Yes, a function can have a limit at a point where it is not defined. This is because a limit only considers the behavior of the function near the specified value, not necessarily at the value itself. As long as the function values approach a specific number as the input values get closer to the specified value, then that number can be considered the limit of the function at that point.