posted by Amber .
1.) Find the equation of the line that is tangent to the graph of y-y^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.
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
y-1 = 3(x-1)
y-1 = 3x-3
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