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Lim t approaches 9 9-t/3-sqrt of t

  • #1
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.... Lim t approaches 9 9-t/3-sqrt of t. First I multiplied the conjugate of 3 + sqrt of t to both numerator and denominator. I got 27+9 sqrt of t - 4t / 9-t .... I plug 9 to t. Denominator cancels out. I got 18 but answer states 6. How?
 

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  • #2
VietDao29
Homework Helper
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.... Lim t approaches 9 9-t/3-sqrt of t. First I multiplied the conjugate of 3 + sqrt of t to both numerator and denominator. I got 27+9 sqrt of t - 4t / 9-t .... I plug 9 to t. Denominator cancels out.
Your numerator after multiplying by [tex]3 + \sqrt{t}[/tex] is incorrect!! You shouldn't expand everything out like that. After multiplying, keep the numerator, and try canceling out instead.

Oh, and by the way, you should learn https://www.physicsforums.com/showthread.php?t=8997" to make your post much clearer. Just take a look at the page I gave. Learning LaTeX isn't very hard, but I assure that it'll give your post a whole new look. Just give it a try. :)
 
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  • #3
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.... Lim t approaches 9 9-t/3-sqrt of t.
At the very least, use enough parentheses to make it clear what you mean. Knowledgeable people would reasonably interpret the expression you wrote as
[tex]9 - \frac{t}{3} - \sqrt{t}[/tex]

Even without using LaTeX you could write your limit like this: lim(t -> 9) (9 - t)/(3 - sqrt(t))
 
  • #4
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Lim t ---> 9

[tex]\frac{(9-t)}{(3-\sqrt{t})}[/tex]
 
  • #5
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Here's how I did it...

[tex]\frac{9-t}{3-\sqrt{t}}[/tex] ([tex]\frac{3+\sqrt{t}}{3+\sqrt{t}}[/tex]) =

[tex]\frac{27+9\sqrt{t}-3t-t\sqrt{t}}{9-t}[/tex]

Plugged in 9 for t...

then I got a different answer than 6.
 
  • #6
eumyang
Homework Helper
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Here's how I did it...

[tex]\frac{9-t}{3-\sqrt{t}}[/tex] ([tex]\frac{3+\sqrt{t}}{3+\sqrt{t}}[/tex]) =

[tex]\frac{27+9\sqrt{t}-3t-t\sqrt{t}}{9-t}[/tex]

Plugged in 9 for t...

then I got a different answer than 6.
No, don't do that. Instead, notice that when you did multiply the denominator by its conjugate you got
[tex](3 - \sqrt{t})(3 + \sqrt{t}) = 9 - t[/tex]
?

What you should do is FACTOR the numerator, which is (sort of) a difference of two squares:
[tex]\displaystyle\lim_{t \rightarrow 9} \frac{9 - t}{3 - \sqrt{t}} = \displaystyle\lim_{t \rightarrow 9} \frac{(3 - \sqrt{t})(3 + \sqrt{t})}{3 - \sqrt{t}}[/tex]

Can you take it from here?
 
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  • #7
Delta2
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Just keep the numerator as (9-t)(3+sqrt(t)) so it will cancel out with the denominator. What is left is 3+sqrt(t).
 
  • #8
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Thanks a lot for your help everyone. It makes a lot of sense now.
 
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  • #9
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Yea I got it

[tex]\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

=3+[tex]\sqrt{t}[/tex]

=6

Thank you. :)
 
  • #10
33,505
5,191
Yea I got it

[tex]\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

= 3+[tex]\sqrt{t}[/tex]

= 6

Thank you. :)
The last two expressions above aren't equal. What you are missing is any indication that you are taking a limit.


What you have above should be written like this.
[tex]\lim_{t \to 9}\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

[tex]=\lim_{t \to 9}3+\sqrt{t}[/tex]

= 6
If you omit the limit symbols, your instructor is likely to deduct points from your work.
 

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