Problem 37 Medium Difficulty

Prove that $ \displaystyle \lim_{x \to a} \sqrt{x} = \sqrt{a} $ if $ a > 0 $.

$ \biggl[Hint: \text{Use $ | \sqrt{x} - \sqrt{a} | = \frac{| x - a |}{\sqrt{x} + \sqrt{a}} $}. \biggr] $

Answer

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Video Transcript

This is problem number thirty seven of the Stuart Calculus eighth edition section two point four. Prove that the limit as expert as a of the function square root of X is equal to the square root of a if a better than zero hint. Use the following identity, and we will see where this comes into play. But let's first recall Thie Absalon. Delta definition would limit for any Absalon is greater than zero. There's a delta graded in zero. You should find a delta greater than zero such that if he absolutely, with the difference between Max and A is this in Delta didn't have some value. The difference between f l is less than epsilon. And we can really write this in terms of what we have in our problem. A bean, eh? In this case, less than daughters. So that is about the same about our function here is going to be squared of X finest Lim score today his Listen, Absalom. And then here we see where the hint will come in because right now we have, ah, suitability of a difference of square roots that we don't really know how to work with. But using this to our advantage, we're going to transform this into ah, or re write the whole thing again. This is Listen, Absalon, but we replaced this, using our hint to make it into this. It's absolutely of the different Kleenex. And A is school out of X plus squared of a which is listen up, Salon. We're gonna make an observation. Ah, we know that the square root of X dysfunction eyes Ah, a positive value. Andi, If we hey simply remove it, we can stay true to the inequality. Ah, and have the comparison to Absalon. Just be the function here divided by the square today on this maintains the inequality. And that's an assumption that we made and such that we can solve for the difference between X and A That's a valley of that is going to be scared of a times Absalon. And now that we made this distinction, we can see the adults that I should be a good choice for. Delta should be that Aah! Delta is equal to escort of eight times Absalon on. We can show directly how this will prove this limit. We're going to take Ah, this exact step here this term this term here It's creative experience saying over score of X, Let's score today. Ah, we see that this it should be less than Epsilon because this is the case on. Then we make the substitution that absolutely of the difference between ixnay over just the skirt of a that still holds true. Ah, this limiting value. We chose to be Delta. So that's sort of a ah long. We will take this term actually which is a squirt of a it's delta squad of eight times up, absolute over this court of aging which we see cancels out to be Absalon on that is consistent with our definition. Therefore, this limit is expert is a of the functions Court of X is equal to the square today.