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

[transgalactic]

<br /> \lim_{x-&gt;3^-} ln^2(9-x^2)=+\infty<br />
the meaning of this ln is e^+infinity=closser to 0

how is that??

by my math e^+infinity is infinity

 

[JG89]

As x approaches three, then ln(9 - x^2) approaches 0 and so the value goes to negative infinity, squaring it gives you positive infinity.

 

[Mark44]

As x approaches three, then ln(9 - x^2) approaches 0 and so the value goes to negative infinity, squaring it gives you positive infinity.

Not quite. As x approaches 3 from below, 9 - x^2 approaches 0, from the positive side, so ln(9 -x^2) approaches negative infinity, so [ln(9 - x^2)]^2 approaches positive infinity.

 

[JG89]

Oops, I meant to say that the value within the ln approaches 0 and so ln approaches negative infinity, which when squared is positive infinity.

 

[Mark44]

Happens to the best of us