Calculus help (limits)? PLEASE

my calc teacher doesn't explain anything so...i'm pretty confused. i found the definition that
lim as x--> 0+ ln x = -infinity
and
lim as x--> +infinity ln x = +infinity.
but i still don't get how to figure out some of the problems even with that.
here are a couple i dont understand plz help!! 10 pts best answer.
1. lim as x---> pi/2 from - side of ln(tan x)
2. lim as x---> +infinity of ln(X^2 - 1) - ln(X + 1)
3. lim as x-->+infinity of ln(3x + 1) - ln(2x^2 + 1)

the rest i've figured out.
please help - my teacher is terrible and i don't understand much :/

1. In the limit, you just sustitute in what the limit of x is into x:
If x=pi/2,
ln(tan x) = ln (tan pi/2)
=ln (infinity)
=infinity.

2. You do the same for the others:
If x=infinity
ln(x^2-1) - ln(x+1) = ln(infinity^2 - 1) - ln(x+1)
=ln(infinity^2) - ln(infinity)
=2ln (infinity) - ln (infinity)
=ln(infinity)
=infinity

In that question, the infinity overpowers the -1 and +1 so they may as well not exist, then you use the rules of logarithms to move the ^2 to the outside.

3. If x=infinity

ln(3x + 1) - ln(2x^2 + 1) = ln ((3x+1)/(2x^2 + 1))
Combine the logarithms (laws of logs).
The +1 on the top and bottom can be ignored again as they don't make any difference when dealing with infinity:

=ln (3x/2x^2)
You can cancel this down:
=ln(3/2x)

Now if x=infinity,

=ln(1/infinity)
=ln(infinity^(-1))
= - ln (infinity)
= - infinity

Sorry this is a bit incoherent.

This is not mathematically correct. Infinity - infinity cannot be simplified to infinity.

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The two formulas come from the graph of ln (x), which goes up forever on the right and down forever on the left, close to zero.
1. lim as x---> pi/2 from - side of ln(tan x)

At pi/2, tanx has an asymptote, going up on the left, and down on the right, so tanx is approaching + inf. As x->( pi/2)-

Now from the second ln formula you have, as u-> + inf, ln(u) -> + inf

So the limit is +inf

2. lim as x---> +infinity of ln(X^2 - 1) - ln(X + 1)

Use properties of limits on these last two.

Lim(x->+ inf) ln [ (x+1)(x-1)/(x+1)]

= lim(x->+ inf) ln (x-1)

so as x-> inf, x-1 -> +inf, so ln(x-1)->+ inf

3. lim as x-->+infinity of ln(3x + 1) - ln(2x^2 + 1)

= lim (x-> + inf) ln [(3x+ 1)/(2x^2+1)]

Now think about horizontal asymptotes: the limit of that fraction is zero, as x-> +inf, because the denominator is higher degree.

So as u->0+ , ln(u)-> -inf

Hoping this helps!