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I am a high school student in Calculus, and we are finishing learning basic limits. I am reviewing for a big test tomorrow, and I could do all of the problems correctly except this one.

I have no idea how to solve the problem this problem correctly. I looked up the answer online, but I can't figure out how they got their answer. All of the online tools show the steps using L'Hospital's rule or derivation, but I haven't learned either yet.

This is the problem:

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

This is the problem that I did incorrectly. I converted the $-1$ to $\frac{\sqrt{1+x}}{\sqrt{1+x}}$, then subtracted the fraction, and multiplied the result by $\frac{1}{x}$ to remove the double division.

$$\large\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}$$

When I substitute $x$, I get $0$, but the answer is $-\frac{1}{2}$. I am doing something simple incorrectly, but I really cannot figure it out.

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As you did, convert to $\dfrac{1-\sqrt{1+x}}{x\sqrt{1+x}}$. (If you substitute, you still get $0$ in the denominator, so there is a little more work to be done.)

Then use the method of "algebraic conjugates":

$$\dfrac{1-\sqrt{1+x}}{x\sqrt{1+x}} \cdot \dfrac {1+\sqrt{1+x}}{1+\sqrt{1+x}} = \dfrac{1-(1+x)}{{x \sqrt{1+x}(1+\sqrt{1+x}})}$$

and simplify, then substitute, to your answer.

If you want to learn the formatting here, it's done in $\LaTeX$, between \$s. To do a fraction, write something like $\frac{\sqrt{2}}{2}$- there are lots of tutorials online.

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  • $\begingroup$ Excellent answer, I forgot to multiply by the conjugate! Thanks, it was really helpful. I'll give LaTeX a look, and accept your answer when the timer is up. $\endgroup$ – Josue Espinosa Sep 16 '14 at 23:38
  • $\begingroup$ @JosueEspinosa codecogs.com/latex/eqneditor.php is helpful for learning $\LaTeX$. $\endgroup$ – Akiva Weinberger Sep 17 '14 at 0:45
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$$\lim_{x\to 0}\frac{\frac{1}{\sqrt{1+x}}-1}{x}=\lim_{x\to 0}\frac{1}{x\sqrt{1+x}}-\frac{1}{x}=$$ $$=\lim_{x\to 0}\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}=\lim_{x\to 0}\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}=$$ $$=\lim_{x\to 0}\frac{1-(1+x)}{x\sqrt{1+x}(1+\sqrt{1+x})}=\lim_{x\to 0}\frac{-x}{x\sqrt{1+x}(1+\sqrt{1+x})}$$ $$=\lim_{x\to 0}\frac{-1}{\sqrt{1+x}(1+\sqrt{1+x})}=-\frac{1}{2}$$

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The formula :$$f(x)= \frac{1-\sqrt{1+x}}{x \sqrt{1+x}}$$is nont able to give the limit because it gives the indeterminate form $\frac00$.

Using the conjugate expression of $1-\sqrt{1+x}$, you get: $$f(x)=\frac{-x}{x \sqrt{1+x}(1+\sqrt{1+x})},$$ then: $$f(x)=\frac{-1}{ \sqrt{1+x}(1+\sqrt{1+x})}$$ and the limit is done.

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$$ \displaylines{ f\left( x \right) = \frac{1}{{\sqrt {1 + x} }} \Rightarrow f'\left( x \right) = \frac{{ - 1}}{{2\left( {1 + x} \right)\sqrt {1 + x} }} \cr f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} \cr = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 + x} }} - 1}}{x} = \frac{{ - 1}}{{2\left( {1 + 0} \right)\sqrt {1 + 0} }} \cr \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 + x} }} - 1}}{x} = - \frac{1}{2} \cr} $$

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