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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 $\sqrt{1+t}+\sqrt{1-t}$. – André Nicolas Aug 28 '11 at 1:55
@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? – anon Aug 28 '11 at 2:33
@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. – André Nicolas Aug 28 '11 at 2:39
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.) – anon Aug 28 '11 at 2:43
@anon: Yes, ping seems to be multiculturally correct, and is not sensitive to accents, or lack of them. – André Nicolas Aug 28 '11 at 2:48
up vote 1 down vote accepted

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/t – user138246 Aug 28 '11 at 2:02
@Jordan Check your algebra. If you show your work above we can help you pinpoint your error. – Bill Dubuque Aug 28 '11 at 2:07
@Bill Dubuque means multiply by a sneaky 1, viz. $\frac{\sqrt{1+t} + \sqrt{1-t}}{\sqrt{1+t} + \sqrt{1-t}}.$ – Robert Haraway Aug 28 '11 at 2:08
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 $0$. – André Nicolas Aug 28 '11 at 2:09
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? – user138246 Aug 28 '11 at 2:09



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



The top is $2t$, so this is


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


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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 $(1+x)^{-\frac{1}{2}} = 1 + \frac{x}{2} + O(x^2)$ – Quixotic Aug 28 '11 at 7:36
leon (and @Tretwick), in what course (formal or otherwise) did you learn about big-O notation? – The Chaz 2.0 Aug 29 '11 at 23:00

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