Using the squeeze theorem to solve a limit

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  • #1
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Homework Statement


I have a limit, which when substitution is used the indeterminate form 0/0 occurs. I've been asked to solve the limit by using the squeeze theorem. I really have no idea where to take this.


Homework Equations


Lim (x[itex]^2[/itex]-9)(x-3[itex]/[/itex]|x-3|)
x[itex]\rightarrow[/itex]3


The Attempt at a Solution


I've tried so many attempts at this problem, posting all my work would be massive.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


I have a limit, which when substitution is used the indeterminate form 0/0 occurs. I've been asked to solve the limit by using the squeeze theorem. I really have no idea where to take this.


Homework Equations


Lim (x[itex]^2[/itex]-9)(x-3[itex]/[/itex]|x-3|)
x[itex]\rightarrow[/itex]3


The Attempt at a Solution


I've tried so many attempts at this problem, posting all my work would be massive.
Are you sure the problem is what you state, which is equivalent to [itex]\displaystyle \lim_{x\to3}\left((x^2-9)\left(x-\frac{3}{|x-3|}\right)\right)\,,[/itex] or is it [itex]\displaystyle \lim_{x\to3}\left((x^2-9)\frac{x-3}{|x-3|}\right)\,?[/itex]
 
  • #3
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It's your second one. I'm a bit confused by the format used on the forum.
 
  • #4
SammyS
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It appears to me that simplifying the expression using algebra works better than using the squeeze theorem, but ...

Certainly, [itex]\displaystyle (x^2-9)\frac{x-3}{|x-3|}\ge0[/itex] for x > 0.

Can you show that [itex]\displaystyle (x^2-9)\frac{x-3}{|x-3|}< 8|x-3|[/itex] for 2 < x < 4 ?
 

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