AP Calc Questions! Find the limit algebaically! HELP ASAP

Solve the following limits algebraically, show all work:

1.) lim as x approaches 0+ of ((x)/(abs.(x))

2.) lim as x approaches 0 of ((x^2-3x)/(sinx))

Please show all work and explain.
Thank you so much for any help you can offer to this struggling Calc student.
God Bless! :)

1.

As x approaches 0 from the right, then x > 0 as it approaches 0
Therefore, as x approaches 0 from the right, | x | = x

lim[x→0⁺] (x / | x |)
= lim[x→0⁺] (x / x)
= lim[x→0⁺] (1)
= 1

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2.

Note that lim[x→0] sinx/x = 1
Therefore, lim[x→0] x/sinx = 1/1 = 1

lim[x→0] ((x²−3x)/sinx)
= lim[x→0] (x(x−3)/sinx)
= lim[x→0] ((x/sinx) (x−3))
= 1 (0−3)
= −3

1) The answer is 1. Thinking of the the graph of x/abs(x), the value can only be 1 of course because it's x/x basically. But the abs value means that when x is negative, then we get -1. So if approaching 0 from the right side, the limit is 1, while approaching 0 from the left side, the limit is -1.

2) Well first thing to notice is plugging in 0, we get 0/0. This is a form of L'hopitals rule, so take the derivative of the top and bottom. So our new form is:
lim as x approaches 0 of ((2x - 3)/(cosx)). Now plug in 0 and you get -3/1, so -3.

Hope this helps!