Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

share|improve this question

7 Answers 7

up vote 2 down vote accepted

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

share|improve this answer
    
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?)

share|improve this answer
    
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 $-1/2$.

share|improve this answer
    
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}

share|improve this answer
    
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}

share|improve this answer

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

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

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

From here for the given limits have:

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.