The discussion revolves around calculating the limit of the expression $$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$. Participants explore various methods to resolve the indeterminate form encountered when substituting $x=0$, including algebraic manipulation and the application of L'Hôpital's rule.
Participants generally agree on the need to manipulate the expression to resolve the indeterminate form, but there are multiple approaches suggested, including algebraic manipulation and L'Hôpital's rule. No consensus is reached on a single method as the best approach.
Participants express uncertainty about the steps involved in manipulating the expression and the application of L'Hôpital's rule. The discussion reflects varying levels of familiarity with these techniques.
This discussion may be useful for students learning about limits, particularly those encountering indeterminate forms and exploring different methods for evaluation.
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 $$\frac{0}{0}$$, 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 $$\sqrt{1+x}-\sqrt{1-x}$$ ?
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:$$1-x$$ ??
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:$$\sqrt{1+x}+\sqrt{1-x}$$ ?
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:
$$1=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$
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:
$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$
wishmaster said:So i multiply all together?
$$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$ ?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:
$$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$
Divide top and bottom with x and Then calculate the limit! Good job!:)wishmaster said:$$\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$ ?