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

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The limit of 1/(x^2 - 9) as x approaches -3 from the left is discussed, with calculations showing that as x gets closer to -3, the function values increase significantly. The denominator approaches zero and remains negative, leading to the conclusion that the limit tends toward positive infinity. Despite some confusion regarding the signs, the consensus is that the limit is indeed positive infinity due to the behavior of the function near this point. The discussion references numerical examples to support this conclusion. Overall, the limit is confirmed to be positive infinity as x approaches -3 from the left.
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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
 
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