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Just another limit problem

  • Thread starter Macleef
  • Start date
30
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1z6ftwo.png
 

Answers and Replies

139
0
You should have 0/0 not 6/0. In any case, try a trick similar to what you did with the radical in the denominator.
 
41
0
Well if you plug the four in initially you get:

2-2/(3-3)=0/0

Your trick did not work nor does using the other radical as Mathdope suggests.

But, never fear, 0/0 limits call for one man.....


L'Hopital!

Since it is indeterminate form, it is eligible for L'Hopital's rule. That should make it work.
 
Dick
Science Advisor
Homework Helper
26,258
618
Well if you plug the four in initially you get:

2-2/(3-3)=0/0

Your trick did not work nor does using the other radical as Mathdope suggests.

But, never fear, 0/0 limits call for one man.....


L'Hopital!

Since it is indeterminate form, it is eligible for L'Hopital's rule. That should make it work.
Multiplying by 2+sqrt(x) does so work. You can then cancel the factor that's going to zero in the numerator and the denominator.
 

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