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 May 10th, 2013, 08:40 PM #1 Newbie   Joined: May 2013 Posts: 9 Thanks: 0 How to solve this limit lim x approches to 0 ?x -?a /x-a = ?
 May 10th, 2013, 10:01 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: How to solve this limit I am assuming you mean: $\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{x-a}$ If this is the case, the try factoring the denominator as the difference of squares.
May 10th, 2013, 11:05 PM   #3
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Re: How to solve this limit

Quote:
 Originally Posted by MarkFL I am assuming you mean: $\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{x-a}$ If this is the case, the try factoring the denominator as the difference of squares.
Wil U Please Tell Me That How To Type That Limit As You Do Above

 May 10th, 2013, 11:19 PM #4 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: How to solve this limit Code: $\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{x-a}$
May 11th, 2013, 06:26 AM   #5
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Re: How to solve this limit

Hello, Bhushan!

Another approach . . .

Quote:
 $\lim_{x\to a}\,\frac{\sqrt{x}\,-\,\sqrt{a}}{x\,-\,a}$

$\text{Multiply by }\frac{\sqrt{x}\,+\,\sqrt{a}}{\sqrt{x}\,+\,\sqrt{a }}$

[color=beige]. . [/color]$\frac{\sqrt{x}\,-\,\sqrt{a}}{x\,-\,a}\,\cdot\frac{\sqrt{x}\,+\,\sqrt{a}}{\sqrt{x}\, +\,\sqrt{a}} \;=\; \frac{x\,-\,a}{(x\,-\,a)(\sqrt{x}\,+\,\sqrt{a})} \;=\;\frac{1}{\sqrt{x}\,+\,\sqrt{a}}$

$\text{Therefore: }\:\lim_{x\to a}\,\frac{1}{\sqrt{x}\,+\,\sqrt{a}} \;=\;\frac{1}{\sqrt{a}\,+\,sqrt{a}} \;=\;\frac{1}{2\sqrt{a}}$

May 11th, 2013, 06:29 AM   #6
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Re: How to solve this limit

Quote:
Originally Posted by soroban
Hello, Bhushan!

Another approach . . .

Quote:
 $\lim_{x\to a}\,\frac{\sqrt{x}\,-\,\sqrt{a}}{x\,-\,a}$

$\text{Multiply by }\frac{\sqrt{x}\,+\,\sqrt{a}}{\sqrt{x}\,+\,\sqrt{a }}$

[color=beige]. . [/color]$\frac{\sqrt{x}\,-\,\sqrt{a}}{x\,-\,a}\,\cdot\frac{\sqrt{x}\,+\,\sqrt{a}}{\sqrt{x}\, +\,\sqrt{a}} \;=\; \frac{x\,-\,a}{(x\,-\,a)(\sqrt{x}\,+\,\sqrt{a})} \;=\;\frac{1}{\sqrt{x}\,+\,\sqrt{a}}$

$\text{Therefore: }\:\lim_{x\to a}\,\frac{1}{\sqrt{x}\,+\,\sqrt{a}} \;=\;\frac{1}{\sqrt{a}\,+\,sqrt{a}} \;=\;\frac{1}{2\sqrt{a}}$

A technical point: $\frac{\sqrt{x}- \sqrt{a}}{x- a}$ is equal to $\frac{1}{\sqrt{x}+ \sqrt{a}}$ for all x except x= a. Fortunately, equality at x= a is not necessary to the limits being the same.

May 13th, 2013, 06:18 PM   #7
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Re: How to solve this limit

Quote:
Originally Posted by HallsofIvy
Quote:
Originally Posted by soroban
Hello, Bhushan!

Another approach . . .

Quote:
 $\lim_{x\to a}\,\frac{\sqrt{x}\,-\,\sqrt{a}}{x\,-\,a}$

$\text{Multiply by }\frac{\sqrt{x}\,+\,\sqrt{a}}{\sqrt{x}\,+\,\sqrt{a }}$

[color=beige]. . [/color]$\frac{\sqrt{x}\,-\,\sqrt{a}}{x\,-\,a}\,\cdot\frac{\sqrt{x}\,+\,\sqrt{a}}{\sqrt{x}\, +\,\sqrt{a}} \;=\; \frac{x\,-\,a}{(x\,-\,a)(\sqrt{x}\,+\,\sqrt{a})} \;=\;\frac{1}{\sqrt{x}\,+\,\sqrt{a}}$

$\text{Therefore: }\:\lim_{x\to a}\,\frac{1}{\sqrt{x}\,+\,\sqrt{a}} \;=\;\frac{1}{\sqrt{a}\,+\,sqrt{a}} \;=\;\frac{1}{2\sqrt{a}}$

A technical point: $\frac{\sqrt{x}- \sqrt{a}}{x- a}$ is equal to $\frac{1}{\sqrt{x}+ \sqrt{a}}$ for all x except x= a. Fortunately, equality at x= a is not necessary to the limits being the same.
[color=#0000FF]Will Anyone Please Tell Me How To Type That Limit As Image Please[/color]

 May 13th, 2013, 07:40 PM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: How to solve this limit Read zaidalyafey's post above. Also, see here.

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