Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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… 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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

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

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.