Why Does the Limit of ln(x²-9) Approach -∞ as x Approaches 3?

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I'm on mobile so I won't use latex.

Lim x -> 3 ln(x^2-9)

My question is why limit approaches - infinite? When I plug in the numbers I get positive infinite.
 
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How did you get positive infinite?? Can you show me what you did?
 
Since I have to figure out x>3, I plugged in the values such as x=4 y=1.945, x=3.5 y=1.178, x=3.9 y=1.826. Whoops I found my mistake haha. I kept plugging in 3.9 and 3.9999 thinking it was closer to 3, when I should had plugged in 3.01 and 3.001. Thank you.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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