Limit as x -> 0 for ((√1+x)-1)/x

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I need to find the limit as x-> 0 for ((√1+x)-1)/x
but I can't figure out how to get rid of x in the denominator or atleast make the limit not equal to 0 in the denominator. I know this is probably going back some, but it's been a little while since I've done this. I've tried about every trick I can think of, but I can't seem to figure it out, but my book seems to think there's an answer.
 
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Are you familiar with l'Hopital's rule?
 
Tide, L'Hopital's rule is "overkill"!

Jeann25, multiply both numerator and denominator by sqrt(1+x)+ 1 to rationalize the numerator.
 
Thank you for the help!
 
HallsofIvy said:
Tide, L'Hopital's rule is "overkill"!
Jeann25, multiply both numerator and denominator by sqrt(1+x)+ 1 to rationalize the numerator.

Halls,

Of course! But I was trying to ascertain whether the assignment related to a lesson on l'Hospital. :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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