1.) Evaluate Lim x->0 (tanx/x)
2.) Find the equations of the lines that pass through the point (0,-2) and are tangent to the graph of y=x^2
3.) Find (d^2*y)/dx^2 and (dy/dx)^2, given that y=7x
1. lim tanx/x , as x ---> 0 (after each line below)
= lim (sinx/cosx)/x
= lim (sinx/x)(1/cox
= lim sinx/x * lim 1/cosx , you should know these
= (1)(1)
= 1
2.
y = x^2
dy/dx = 2x
Let the point of contact be (a,b)
we know b = a^2
and the slope of the tangent is 2a
but using the good ol' grade 9 way of finding the slope
slope of tangent = (b+2)/a
thus (b+2)/a = 2a
2a^2 = b+2 , but b = a^2
2a^2 = a^2 + 2
a^2 = 2
a = ±√2
I will find one tangent, you find the other in the same way
if a=√2, then b = 2
so one point of contact is (√2,2)
at that point slope = 2√2
equation:
y-2 = 2√2(x-√2)
y = 2√2 x - 2√2 + 2
3.
I will interpret (d^2*y)/dx^2 as the second derivative and (dy/dx)^2 as the square of the first derivative
y = 7x
y' = 7 , so (dy/dx)^2 = 49
y '' = 0