$\lim_{t\to 0} \frac{(1+t)^{1/2} - (1-t)^{1/2}}{t}$

How can I find the limit of this?

Should I use the conjugate pair?

It doesn't seem logical that this has a limit, I started by trying to simplifying the function but I don't think you can simplify this any further. $$\lim_{t\to 0} \frac{(1+t)^{1/2} - (1-t)^{1/2}}{t}$$

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thank you thomas. i need to fix the (1-t) it should be (1-t)^1/2 as well – Miguel Jan 30 '13 at 18:13
You can use LaTeX (MathJax) to format your question. See meta.math.stackexchange.com/questions/5020/… for more on this. – Thomas Jan 30 '13 at 18:14
Yes, multiplying by the conjugate is a good idea. And it works. – 1015 Jan 30 '13 at 18:14
Have you learned derivatives? – Jonas Meyer Jan 30 '13 at 18:14
@Miguel: Just asking because rewriting it as $\frac{(1+t)^{1/2}-1}{t}-\frac{(1-t)^{1/2}-1}{t}$ would be a way to quickly see what the limit is if you already know enough about derivatives. Regardless, maybe you'll find these two limits a little easier than the original. – Jonas Meyer Jan 30 '13 at 18:19

Hint (as mentioned in the comments). By multiplying by the conjugate you get $$\frac{[(1+t)^{1/2} - (1-t)^{1/2}]}{t}\cdot\frac{[(1+t)^{1/2} + (1-t)^{1/2}]}{[(1+t)^{1/2} + (1-t)^{1/2}]} = \frac{1+t - (1-t)}{t[(1+t)^{1/2} + (1-t)^{1/2}]}$$

Now try to simplify this expression a bit and then look at $t\to 0$ again.

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thank you thomas on both count for teaching me how to properly put my math values into this forums, fixing my question, and on top of that proposing answer im forever in your debt. Thank you. – Miguel Jan 30 '13 at 18:45
@Miguel: I am glad to help. – Thomas Jan 30 '13 at 18:47

$$\lim_{t\to 0} \frac{\sqrt{1+t} - \sqrt{1-t}}{t}=\lim_{t\to 0} \frac{\left(\sqrt{1+t} - \sqrt{1-t}\right)\left(\sqrt{1+t} + \sqrt{1-t}\right)}{\left(\sqrt{1+t} + \sqrt{1-t}\right)t}= \\ =\lim_{t\to 0} \frac{{1+t} - {(1-t)}}{\left(\sqrt{1+t} + \sqrt{1-t}\right)t}=\lim_{t\to 0} \frac{2t}{\left(\sqrt{1+t} + \sqrt{1-t}\right)t} = \\ =\lim_{t\to 0} \frac{2}{\sqrt{1+t} + \sqrt{1-t}}=1$$

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$$\lim_{t\to 0} \frac{(1+t)^{1/2} - (1-t)^{1/2}}{t} = \lim_{t\to 0} \frac{((1+t)^{1/2} - (1 - t)^{1/2})((1+t)^{1/2}+(1-t)^{1/2})}{t((1+t)^{1/2} + (1-t)^{1/2})}$$ $$= \lim_{t\to 0} \frac{1 + t - (1 - t)}{t((1+t)^{1/2} + (1-t)^{1/2})} =\lim_{t\to 0} \frac{2t}{t((1+t)^{1/2} + (1-t)^{1/2})}$$

$$=\lim_{t\to 0}\frac{2}{((1+t)^{1/2} + (1-t)^{1/2})} = \frac{2}{2} = 1$$

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