Limit problem using l'hopital's rule

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  • #1
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Homework Statement



[tex]\lim_{x\rightarrow2}\frac{x\sqrt{x-1}-2}{x-2}[/tex]


Homework Equations



calculate the limit

The Attempt at a Solution



When I apply l'hopital's i have to do derivative of the top over derivative of the bottom right?
 
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Answers and Replies

  • #2
SammyS
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Homework Statement



[tex]\lim_{x\rightarrow2}\frac{x\sqrt{x-1}-2}{x-2}[/tex]


Homework Equations



calculate the limit

The Attempt at a Solution



When I apply l'hopital's i have to do derivative of the top over derivative of the bottom right?
Yes .
 
  • #3
HallsofIvy
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You could use L'Hopital but I would consider that "overkill". Instead, rationalize the numerator and factor:

[tex]\frac{x\sqrt{x- 1}- 2}{x- 2}\frac{x\sqrt{x- 1}+ 2}{x\sqrt{x- 1}+ 2}= \frac{x^2(x- 1)- 4}{(x- 2)(x\sqrt{x- 1}+ 2)}= \frac{x^3- x^2- 4}{(x- 1)(x\sqrt{x- 1}+ 2)}[/tex]
[tex]\frac{(x- 2)(x^2+ x+ 2)}{(x- 2)(x\sqrt{x- 1}- 2)}= \frac{x^2+ x+ 2}{x\sqrt{x- 1}+ 2}[/tex]

Now, let x= 2.

(We knew it would factor that way precisely because x= 2 made both numerator and denominator equal to 0.)
 
  • #4
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Another alternative:


[tex]Let \ x \ =\ y^2 + 1.[/tex]



[tex]As \ \ x\to 2, \ \ y \to 1.[/tex]


[tex]\lim_{y \to 1} \dfrac{(y^2 + 1)\sqrt{y^2} - 2}{(y^2 + 1) - 2} \ =[/tex]

[tex]\lim_{y \to 1}\dfrac{|y|(y^2 + 1) - 2}{y^2 - 1} \ = \ [/tex]

[tex]\lim_{y \to 1}\dfrac{y(y^2 + 1) - 2}{y^2 - 1} \ ** \ = \ [/tex]

[tex]\lim_{y \to 1}\dfrac{y^3 + y - 2}{y^2 - 1} \ = \ [/tex]

[tex]\lim_{y \to 1}\dfrac{(y - 1)(y^2 + y + 2)}{(y - 1)(y + 1)} \ = \ [/tex]

[tex]\lim_{y \to 1}\dfrac{y^2 + y + 2}{y + 1}[/tex]


Continue . . .





[tex]** \ Note: \ \ \ As \ \ y \ \ is \ \ positive \ \ near \ \ 1, \ \ |y| = y.[/tex]
 
  • #5
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I used L'Hopital's rule, and I think it's actually simpler in this case. The derivative of the denominator is 1, the derivative of the constant 2 is 0 so ...
 
  • #6
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I used L'Hopital's rule,
and I think it's actually simpler in this case.

The derivative of the denominator is 1, [tex] \ \ \ \ Yes, \ \ so \ \ evidently \ \ you're \ \ done \ \ discussing \ \ the \ \ denominator.[/tex]


the derivative of the constant 2 is 0 [tex] \ \ \ \ No, \ \ if \ \ you \ \ are \ \ now \ \ discussing \ \ the \ \ numerator, [/tex][tex] you \ \ are \ \ to \ \ be \ \ taking \ \ the \ \ derivative \ \ of \ \ \ x\sqrt{x - 1} - 2. [/tex]
[tex] And \ \ that \ \ derivative\ \ is \ \ not \ \ equal \ \ to \ \ zero. [/tex]



so ...

....
 
  • #7
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I meant that the derivative of the constant -2 (which appears in the numerator) is 0, so we just have to use the product rule to take the derivative of x*sqrt(x-1). The result,

√(x-1) + (x/2)*((√(x-1))^(-1))

... sorry I don't have practice entering these equations on this system ... but now you can just sub in x = 2 and the limit falls right out.

√(2-1)+(2/2)*((√(2-1)^(-1))

√1 + (1)*(1/√1)= 2

what could be simpler?

I hope I put in the right number of parens.
 
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  • #8
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Yes I solved it guys thanks for the help. By the way the problem stated on solving it using l'hopital's rule
 

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