Limit of (sqrt(x) - 2)(x-4) as x approaches 4 from the left

  • Thread starter razored
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  • #1
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Homework Statement


[tex]\lim_{x\rightarrow 4^{-}}\frac{\sqrt{x}-2}{x-4}[/tex]
Evaluate

The Attempt at a Solution


Well, I used a calculator to substitute arbitrary values to see what it appears to approach.

input | output
2 .293
3 .268
3.5 .258
3.9 .252

So I concluded that the the limiting value is in fact .25.. and I was wrong. How do I evaluate this?

NOTE THAT WE ARE APPROACHING FROM THE LEFT!
 
Last edited:

Answers and Replies

  • #2
Defennder
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Use L'Hospital's rule.
 
  • #3
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Thank you, I got it.
 
  • #4
HallsofIvy
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A fairly standard way of handling a problem like that, and simpler than apply L'Hopital's rule is to rationalize the numerator. Multiplying both numerator and denominator by [itex]\sqrt{x}+ 2[/itex] you get
[tex]\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}= \frac{x- 4}{(x-4)(\sqrt{x}+ 2)}[/tex]
which, for x> 4, is
[tex]\frac{1}{\sqrt{x}+ 2}[/tex]
Of course, the limit of that, as x goes to 2, is the same as the limit of the original problem.
 
  • #5
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A fairly standard way of handling a problem like that, and simpler than apply L'Hopital's rule is to rationalize the numerator. Multiplying both numerator and denominator by [itex]\sqrt{x}+ 2[/itex] you get
[tex]\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}= \frac{x- 4}{(x-4)(\sqrt{x}+ 2)}[/tex]
which, for x> 4, is
[tex]\frac{1}{\sqrt{x}+ 2}[/tex]
Of course, the limit of that, as x goes to 2, is the same as the limit of the original problem.
Yeah, that is what I did. I'm not up to the rule yet. Thanks again halls of ivy.
 

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