# rational limit problem

I am teaching myself calculus and have run into a limit problem that I do not understand how to solve.

$$\lim_{x\to 0}\left(\frac{1}{x\sqrt{1 + x}}-\frac1x\right)$$

All I am supposed to know at this point about calculus is the 11 limit laws. I think that I am just stuck on the algebra however. I do not know how to rid the denominators of the x's so it is not dividing by zero.

I have tried two ways to solve this problem as they are all I know.

1) obtaining a common denominator

$$lim_{x\to 0}\left({x-x\sqrt{1+x}\over x^2\sqrt{1+x}}\right)$$ $$lim_{x\to 0}\left({1-\sqrt{1+x}\over x\sqrt{1+x}}\right)$$

2) Multiplying to get rid of the denominators square root followed by a common denominator.

$$\lim_{x\to 0}\left({\sqrt{1 + x}\over x(1+x)}-\frac1x\right)$$ $$\lim_{x\to 0}\left({\sqrt{1 + x}-(1+x)\over x(1+x)}\right)$$

Please help me understand what I am missing. Any general principles for understanding/handling problems of this nature would be greatly appreciated.

p.p.s. I don't know if its a good idea to source where i got the problem from, but the book I am using is Single Variable Calculus Early Transcendentals 6E by James Stewart. The problem is in section 2.3, #29.

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You've done very well thus far, but there's one other trick you might want to try...

Continuing from your first approach, multiply above and below by the conjugate of the numerator. That is:

\begin{align} \lim_{x\to 0}\left(\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}\right) &=\lim_{x\to 0}\left(\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}\cdot\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}\right) \\ &=\lim_{x\to 0}\left(\frac{1-(1+x)}{\color{red}{x}\sqrt{1+x}(1+\sqrt{1+x})}\right) \\ \end{align}

Hint: with just minor simplification, you should be able to cancel out that troublesome $\color{red}{x}$.

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ah the conjugate! I even thought about doing that but for some reason I thought it was only useful for the denominator. I probably should have tried it anyway. Thanks for the help! – ludragon Jan 31 '14 at 3:08

$$L=\lim_{x\to0}\frac{\dfrac1{\sqrt{1+x}}-1}x=\lim_{x\to0}\frac{(1+x)^{^{-\frac12}}-1}x\quad\overset{\text{l'Hopital}}{=}\quad\lim_{x\to0}\frac{-\frac12(1+x)^{^{-\frac32}}-0}1=-\frac12$$

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I don't know about this rule yet but im sure it will come in handy when I learn about it :) – ludragon Jan 31 '14 at 3:09