1. ## Improper Integration

I know this is correct, but I'm confused about one thing. When I take the limit as t approaches 4 from the left, do I approach 4 from the left on $\displaystyle \frac{1}{x^2 - 3x - 4}$ or on $\displaystyle ln|x-4|$?

2. If you approached on $\displaystyle \frac{1}{x-4}$, the over all limit would not exist.

3. Originally Posted by dwsmith
If you approached on $\displaystyle \frac{1}{x-4}$, the over all limit would not exist.
Aren't we approaching $\displaystyle ln|x-4|$?

4. Originally Posted by temaire
Aren't we approaching
Originally Posted by temaire
$\displaystyle ln|x-4|$?

That limit doesn't exist. It only exist on the original equation. You can't evaluate this
$\displaystyle ln|x-4|$ at 4 since one side is $\displaystyle +\infty$ and the other side is $\displaystyle -\infty$

5. Originally Posted by dwsmith
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That limit doesn't exist. It only exist on the original equation. You can't evaluate this
$\displaystyle ln|x-4|$ at 4 since one side is $\displaystyle +\infty$ and the other side is $\displaystyle -\infty$
Actually, aren't both sides negative? This is the graph from x=3 to x=5.

6. I understand now. You can approach ln|x-4| from the left.

7. Originally Posted by temaire
Actually, aren't both sides negative? This is the graph from x=3 to x=5.

That is what I was telling you. That limit exist at 4 and the ln function you were asking about DNE