Limit of Rational Function....3

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  • #1
nycmathdad
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Find the limit of 1/(x^2 - 9) as x tends to -3 from the left side.

Approaching -3 from the left means that the values of x must be slightly less than -3.

I created a table for x and f(x).

x...(-4.5)...(-4)...(-3.5)
f(x)... 0.088...0.142...…...0.3076

I can see that f(x) is getting larger and larger and possibly without bound.

I say the limit is positive infinity.

Yes?
 

Answers and Replies

  • #2
jonah1
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Problem 1.5.33.
Odd numbered.
Look up the answer.
 
  • #3
HOI
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For x close to 3 and less than 3, the denominator, x^2- 9, is close to 0 and negative while the numerator, 1, is positive. Therefore?
 
  • #4
nycmathdad
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For x close to 3 and less than 3, the denominator, x^2- 9, is close to 0 and negative while the numerator, 1, is positive. Therefore?

Thus, the limit is positive infinity.
 
  • #5
HOI
923
2
Because the value is negative the limit is positive? That is what you are saying!

If x= 2.9, x^2- 9= -0.59, 1/(x^2- 9)= -1.69491525.
If x= 2.99, x^2- 9= -0.0599. 1/(x^2- 9)= -16.694408.
If x= 2.999, x^2- 9= -166.944.

That is NOT going to be positive!
 
  • #6
jonah1
108
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Beer soaked ramblings follow.
Find the limit of 1/(x^2 - 9) as x tends to -3 from the left side.

Approaching -3 from the left means that the values of x must be slightly less than -3.

I created a table for x and f(x).

x...(-4.5)...(-4)...(-3.5)
f(x)... 0.088...0.142...…...0.3076

I can see that f(x) is getting larger and larger and possibly without bound.

I say the limit is positive infinity.

Yes?
Because the value is negative the limit is positive? That is what you are saying!

If x= 2.9, x^2- 9= -0.59, 1/(x^2- 9)= -1.69491525.
If x= 2.99, x^2- 9= -0.0599. 1/(x^2- 9)= -16.694408.
If x= 2.999, x^2- 9= -166.944.

That is NOT going to be positive!
Country Boy, in case you haven't noticed it yet, nycmathdad has been banned already.
Also, nycmathdad's conjecture about the function's behavior as x approaches −3 from the left (not +3 from the leftt as you seem to have misread), that is about the ratio 1/(x^2 − 9) becoming unbounded in the positive direction is indeed correct.
https://www.desmos.com/calculator/kh87f1cusv
 
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