Limits problems brief calculus help please

I'm stuck on those problems.. I can't seem to find out how to solve it.. My professor is korean and he has a strong accent I can't really understand the way he explains :( .

lim (x+h)^2-(x^)/(h)
h>0
and
lim (x+h)^3-(x)^3/(h)
h>0

lim (3x)/(7x-1)
x>infinite

if u guys know a good website that'll help me out please send it to me ... Thank you

on number one i assume its:

lim h-->0 (x+h)^2 - x^2 / h

Expand (x+h)^2

lim h--> 0 [(x^2+2xh+h^2) - x^2] / h

lim h-->0 (2xh+h^2) / h

lim h--> 0 h(2x + h) / h
lim h-->0 (2x+h) / 1 = 2x

equation was originally x^2---- derivative of x^2 is 2x... so nice :)

For the second one do the exact same thing...

lim h--> 0 (x^3 + 3x^2h + 3xh^2 + h^3) - x^3 / h

lim h--> 0 (h(3x^2 + 3xh + h^2) / h)

lim h--> 0 (3x^2 + 3xh + h^2) = 3x^2

d/dx(x^3) = 3x^2

lim x--> ∞ (3x/7x-1)

divide everything by highest degree of x in denominator:
lim x--> ∞ (3x/7x-1) * (1/x)/(1/x)

lim x--> ∞ 3 / (7 - (1/x))

as x--> ∞ 1/x ---> 0

3/ (7-0) = 3/7

((x + h)^2 - x^2) / h =>
(x^2 + 2hx + h^2 - x^2) / h =>
(2hx + h^2) / h =>
2x + h
h goes to 0
2x + 0 =>
2x



((x + h)^3 - x^3) / h =>
(x^3 + 3x^2 * h + 3x * h^2 + h^3 - x^3) / h =>
(3x^2 * h + 3x * h^2 + h^3) / h =>
3x^2 + 3xh + h^2

h goes to 0

3x^2 + 3x * 0 + 0^2 =>
3x^2



(3x) / (7x - 1)

u = 7x - 1
u + 1 = 7x
(1/7) * (u + 1) = x

3 * (1/7) * (u + 1) / (u) =>
(3/7) * u / u + (3/7) * 1/u =>
(3/7) * 1 + (3/7) * 1/u =>
(3/7) * (1 + 1/u) =>
(3/7) * (1 + 1/(7x - 1))

x goes to infinity

(3/7) * (1 + 1/(inf)) =>
(3/7) * (1 + 0) =>
(3/7) * 1 =>
3/7