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Evaluate each limit (if it exists). Use $L$ Hospital's rule (if appropriate).$$\lim _{x \rightarrow 0} \frac{\sqrt{1-x}-\sqrt{1+x}}{x}$$

-1

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 7

L'Hospital's Rule

Derivatives

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All right. We are doing the limit as X approaches zero and I know it's a square root, but I'm going to rewrite as till one half power because it's gonna be easier when I do the derivative of that. Um And same thing with the one plus X. Very right. That is to the one half power all over X. So before getting to the derivative, you have to make sure that when you plug in zero direct substitution, then you get uh a form that lets you do low petals rule. So one minus zero is one. The square to one is still one. The same thing with one plus one is sorry, one plus zero with direct substitution is one. The square to one is still one. So we do get zero on top. And then the denominator obviously with direct substitution effect zero. That's all we have. So the indeterminant form of 00 does apply. Um to let us use low petals rule. So what we can do then is we can take the limit as X approaches zero. And the derivative of the top is where I'm going to use the chain rule. Bring that one half in front. One minus X. Is now to the negative one half power uh times the derivative of the inside which is negative one. Same thing with the next one. One half in front one plus X. To the negative one half power times the derivative of the inside is just one all over. The derivative of the bottom is one. So just a reminder, what this actually means with direct substitution is we have we were equals negative one half. Um One minus zero is one. And what's nice about this is one to any power is still one. Uh Because technically it's in the denominator is the square root. I don't know if that makes sense. Same thing with the next piece is one plus zero is still one and one to any power still one. Um I know it's in the denominator because of the negative exponents, the square to one is still one. So we're left with one half minus one half, which is just negative two halves or negative one. Which is your correct answer. That was the original limit using low petals rule.

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