Is the Limit of 1/(x^2 - 9) as x Approaches -3 from the Left Positive Infinity?

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Discussion Overview

The discussion revolves around the limit of the function 1/(x^2 - 9) as x approaches -3 from the left. Participants explore the behavior of the function near this point, considering both numerical evidence and theoretical reasoning.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that as x approaches -3 from the left, the function values increase without bound, proposing that the limit is positive infinity.
  • Another participant questions the reasoning, providing specific values for x close to -3 and arguing that the function yields negative results, thus challenging the claim of the limit being positive infinity.
  • Some participants reiterate the behavior of the denominator being close to zero and negative, while the numerator remains positive, leading to the conclusion that the limit should be positive infinity.
  • A later reply emphasizes that the conjecture about the function's behavior as x approaches -3 from the left is correct, supporting the idea of the limit being unbounded in the positive direction.

Areas of Agreement / Disagreement

Participants express disagreement regarding the limit's value, with some asserting it approaches positive infinity and others providing counterexamples that suggest it does not. The discussion remains unresolved.

Contextual Notes

There are conflicting interpretations of the function's behavior near the limit, with some participants relying on numerical examples while others focus on theoretical reasoning. The discussion highlights the complexity of approaching limits in this context.

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?
 
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Problem 1.5.33.
Odd numbered.
Look up the answer.
 
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?
 
Country Boy said:
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.
 
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!
 
Beer soaked ramblings follow.
Country Boy said:
nycmathdad said:
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
 
Last edited:

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