Calculus
posted by Amber .
1.) Find the equation of the line that is tangent to the graph of yy^3 at x=1.
2.) lim x>0 ((sin x*cos 2x)/3x)
3.) Show, using the squeeze theorem, that the limit of Xe^(sin 1/x) as x>0 is 0.

Typo: 1.) Find the equation of the line that is tangent to the graph of y=x^3 at x=1.

1.
y = x^3 , at x = 1, y=1, so we have the point(1,1)
dy/dx = 3x^2 , which is 3, when x = 1
y1 = 3(x1)
y1 = 3x3
3x  y = 2 or y = 3x  2
2. Lim sinxcos 2x/(3x)
= lim sinx/x * cos 2x/3, as x > 0
= (1)*(1/3) = 1/3 
Since (sin 1/x) <= 1
e^(sin 1/x) <= e
so x e^sin(1/x) <= x * e > 0 as x>0