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Find $$ \lim_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}. $$

I can't think of how to start this or what to do at all. Anything I try just doesn't change the function.

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    I expect you are trying to find the limit as $t$ approaches $0$ of your expression. Essentially the same trick as before: multiply "top" and "bottom" by $\sqrt{1+t}+\sqrt{1-t}$.2011-08-28
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    @André: After seeing Bill post a hint (as he usually does) I was feeling iffy about essentially doing OPs homework for them. (I didn't even see the error until you pointed it out......) Also, do you get pinged even if one doesn't use the correct "e" in your name?2011-08-28
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    @anon: I think the OP needs to get some sort of good start, since algebra the OP once knew, and probably got an A on, is now partially forgotten.2011-08-28
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    lol. Jordan: Of course I'm not saying you should not do your homework! I'm saying it's better for answerers to teach so that question-askers understand how to do problems themselves, which I don't know if I was doing effectively. (The fact you have been doing this for longer than you care to bolsters my point that askers would benefit more from a real understanding than simply getting something they can write down.)2011-08-28
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    @anon: Yes, ping seems to be multiculturally correct, and is not sensitive to accents, or lack of them.2011-08-28
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    Here's [another example,](http://math.stackexchange.com/questions/60792/x-frac-b-pm-sqrtb2-4ac2a-show-that-x-c-b-when-a-0/60799#60799) of **rationalizing the numerator** namely, specializing the quadratic formula when $\: a = 0\:.\:$2011-08-30

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HINT $\ $ Use the same method in your prior question, i.e. rationalize the numerator by multiplying both the numerator and denominator by the numerator's conjugate $\rm\:\sqrt{1+t}+\sqrt{1-t}\:.$ Then the numerator becomes $\rm\:(1+t)-(1-t) = 2\:t,\:$ which cancels with the denominator $\rm\:t\:,\:$ so $\rm\:\ldots$

More generally, using the same notation and method as in your prior question, if $\rm\:f_0 = g_0\:$ then

$$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{g(x)}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-g(x)}{x\ (\sqrt{f(x)}+\sqrt{g(x)})}\ =\ \dfrac{f_1-g_1}{\sqrt{f_0}+\sqrt{g_0}}$$

In your case $\rm\: f_0 = 1 = g_0,\ \ f_1 = 1,\ g_1 = -1\:,\ $ so the limit $\: =\: (1- (-1))/(1+1)\: =\: 1\:.$

Note again, as in your prior questions, rationalizing the numerator permits us to cancel the common factor at the heart of the indeterminacy - thus removing the apparent singularity.

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    I am left with 2/t2011-08-28
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    @Jordan Check your algebra. If you show your work above we can help you pinpoint your error.2011-08-28
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    @Bill Dubuque means multiply by a sneaky 1, viz. $\frac{\sqrt{1+t} + \sqrt{1-t}}{\sqrt{1+t} + \sqrt{1-t}}.$2011-08-28
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    You are supposed to multiply "top" and "bottom" by $\sqrt{1+t}+\sqrt{1-t}$. If you do the calculation carefully, watching out for treacherous minus signs, the top should simplify to $2t$. The bottom should be $t(\sqrt{1+t}+\sqrt{1-t})$. When $t \ne 0$, can cancel $t$ from top and bottom. Now top is $2$, bottom is $\sqrt{1+t}+\sqrt{1-t}$. Finally, let $t$ approach $0$.2011-08-28
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    I am going to make the 1+t x and the 1-t y to make it easier to type. I got 1+t +xy - xy -1-t which gives me 0 correct?2011-08-28
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    @Jordan You need to multiply *both* numerator and denominator to obtain an equivalent fraction - just as in the prior question.2011-08-28
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    I tried again and got 2t/(t sqrt 1 + t -sqrt 1 -t)2011-08-28
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    @Jordan: it is much easier to read if you set your math in $\LaTeX$. There are tutorials on the web, here you enclose the $\LaTeX$ in dollar signs. If you right click on any expression and choose Show Source you can see how it was done. If you're not going to do that, please use parentheses: 2t/(t( sqrt (1 + t) -sqrt (1 -t))). Otherwise we won't understand what you mean easily.2011-08-28
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    That is what I did which leaves me with 2/$(\sqrt{1+t}+\sqrt{1-t})$2011-08-28
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    @Jordan That's correct, now evaluate it at $\:t = 0$.2011-08-28
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    Got it, didnt realize that it would be 1+1.2011-08-28
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$$\frac{\sqrt{1+t}-\sqrt{1-t}}{t}$$

$$=\frac{\sqrt{1+t}-\sqrt{1-t}}{t}\cdot\frac{(\sqrt{1+t}+\sqrt{1-t})}{(\sqrt{1+t}+\sqrt{1-t})}$$

Note $(a-b)(a+b)=a^2-b^2$, so this is

$$\frac{(\sqrt{1+t})^2-(\sqrt{1-t})^2}{t(\sqrt{1+t}+\sqrt{1-t})}$$

$$=\frac{(1+t)-(1-t)}{t(\sqrt{1+t}+\sqrt{1-t})}$$

The top is $2t$, so this is

$$\frac{2}{\sqrt{1+t}+\sqrt{1-t}}.$$

To find this as $t\to0$ just plug in $t=0$, which gives

$$\frac{2}{\sqrt{1}+\sqrt{1}}=1.$$

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This is probably not what you are expected to do (you probably are not supposed to know/use Taylor series at this point), but for the sake of information: this kind of limits are more easily solved with Taylor expansions. Knowing that around zero $\sqrt{1+x} = 1 + \frac{x}{2} + O(x^2)$, the result comes immediately.

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    Or Binomial theorem for $(1+x)^{-\frac{1}{2}} = 1 + \frac{x}{2} + O(x^2)$2011-08-28
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    leon (and @Tretwick), in what course (formal or otherwise) did you learn about big-O notation?2011-08-29