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

[nycmathdad]

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?

 

[jonah1]

Problem 1.5.33.
Odd numbered.
Look up the answer.

 

[HOI]

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?

 

[nycmathdad]

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.

 

[HOI]

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!

 

[jonah1]

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