# Finding the limit

I need to find the limit of this problem. I pretty much know you have to multiply by the conjugate but I get lost after I do that.

$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$$

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You don't have to multiply by a conjugate. Hint: $1-x=(1-\sqrt{x})(1+\sqrt{x})$.

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Actually... I messed up the problem. See the OP – user69 Jun 20 '12 at 5:02
@Jamie: The only thing different is the sign, so all I had to do was switch the terms around! – anon Jun 20 '12 at 5:10

You need to find

$$\lim\limits_{x \to 1}\frac{\frac 1 {\sqrt x}-1}{1-x}$$

This is

$$\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x }}\frac{{1 - \sqrt x }}{{1-x}}$$

Can you move on with that? (With anon's hint maybe?)

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yes, thank you! – user69 Jun 20 '12 at 5:21

$$\begin{eqnarray} -\lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{1-x} &=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\\ &=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\frac{1/\sqrt{x}+1}{1/\sqrt{x}+1}\\ &=& \lim\limits_{x\to 1} \frac{1/x - 1}{(x-1)(1/\sqrt{x}+1)}\\ &=& \lim\limits_{x\to 1} \frac{1}{1/\sqrt{x}+1}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}\\ &=& \frac{1}{2}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1} \end{eqnarray}$$ This limit can be evaluated by noting that substituting $1/x$ for $x$ gives one over the limit, but should give the same value since $x\to 1$ is the same as $1/x\to 1$, thus the limit $L$ satisfies $L=1/L$, so $L=1$. This gives us a final answer of $-1/2$.

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I apologize, and thank you for your answer but it was actually 1-x, not x-1! – user69 Jun 20 '12 at 5:08
@Jamie I modified the answer to adapt it, noting that $1-x=-(x-1)$. – Alex Becker Jun 20 '12 at 5:13

\begin{align} \lim_{x \to 1} \dfrac{1/\sqrt{x}-1}{1-x} & = \lim_{x \to 1} \dfrac1{\sqrt{x}}\dfrac{1-\sqrt{x}}{1 - (\sqrt{x})^2} = \lim_{x \to 1} \dfrac1{\sqrt{x}}\dfrac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})}\\ & = \lim_{x \to 1} \dfrac1{\sqrt{x}}\dfrac1{(\sqrt{x}+1)} = \dfrac12 \end{align}

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It should be $\frac{1}{2}$ not $\frac{-1}{2}$, – Joe Jun 20 '12 at 5:15
@JoeL. The OP changed the sign in the edit. – user17762 Jun 20 '12 at 5:16
Ah, you're right, my bad. – Joe Jun 20 '12 at 5:17

I'd opt for L'Hôpital's rule.

\begin{align} &L = \lim_{x \to 1} \dfrac{\dfrac{1}{\sqrt{x}}-1}{1-x}\\ &L = \lim_{x \to 1} \dfrac{1}{-2x^{\frac{3}{2}}} \cdot -1\\ &L = \lim_{x \to 1} \dfrac{1}{2x^\frac{3}{2}}\\ &L = \dfrac{1}{2}\\ \end{align}

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Let's solve it elementarily:

$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}=\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x} \cdot \frac{(1 / \sqrt{x}) + 1}{(1 / \sqrt{x}) + 1}=\lim\limits_{x\to 1}\frac{1-x}{2x(1-x)}=\frac{1}{2}.$$

Q.E.D.

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$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$.

We substition $\sqrt {x}=t$, hance we:

If $x\longrightarrow 1\Rightarrow t\longrightarrow 1.$

From here for the given limits have:

$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$=$\lim\limits_{t\to 1} \frac{\frac{1}{t} - 1}{1-t^2}$=$\lim\limits_{t\to 1} \frac{\frac{1-t}{t}}{1-t^2}$=$\lim\limits_{t\to 1} \frac{1-t}{t(1-t)(1+t)}$=$\lim\limits_{t\to 1} \frac{1}{t(t+1)}$=$\frac{1}{2}$

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