How to do C and D (calc 2)

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[languid geyser]

Imagine the graph of $\int_{0}^{1}\frac{1}{x}dx$ for C

[craggy seal]

+C Forgetter

[languid geyser]

If you solve this improper integral you'll use a limit and see that it doesn't converge

yes but the limit as x approaches 0 of lnx is not undefined it diverges to negative infinity

It is but we aren't finding ln0. We're finding $\lim_{a\to0}ln(1)-ln(a)$

[craggy seal]

+C Forgetter

[languid geyser]

The limit of ln(a) as 'a' approaches 0 is not equal to ln(0)

Here is a graph of lnx. As x gets closer and closer to 0, lnx gets closer and closer to negative infinity. A limit asks "As x gets closer to 0, what does lnx get closer to?". It does not matter what the value of the function is at 0, all that matters is what it's approaching.

Yes $\lim_{a\to0}ln(1)-ln(a)$ diverges to infinity

[craggy seal]

+C Forgetter

[languid geyser]

But it can be somewhat confusing using = sign with limits. With limits we don't actually care what the value is at the point it approaches. For a convergent series we can use f(x) = 1.

I'd start with knowing $0\leq f(x)\leq \frac{1}{x}$

[craggy seal]

+C Forgetter

[languid geyser]

Then plug in values for 0 and 1/x

Solve and see which converge and diverge. Afterwards we can use the fact that if a function is smaller than something that converges it also converges and if something is larger than something that diverges, it also diverges.