x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)
So, for all x≠1,
f(x) = (x-2)(x-3)
as x->1, f(x)->2 since both factors are negative
we need to show that for every ϵ>0 there is a δ such that
f(x+δ)-2 < ϵ
we can dispense with the absolute value stuff, since f(x) > 0 and we are taking the upper limit. So, we just need to show that we can solve for δ, no matter which small ϵ we choose.
((x+δ)-2)((x+δ)-3)-2 < ϵ
That's just a simple quadratic, which will have two real roots.
consider the statement
lim x³-6x²+11x-6 / x-1 = 2
x->1
Using the definition of the limit, state what must be true for the above limit to hold, that is, for
every ..., there is ..., so that.... Use a specific function and limit not just f and L.
� Verify the limit is true by finding ä as an expression of ϵ.
� Draw a picture illustrating the relation between ϵ, ä and the function.
i am at hulk422 at g mail . com
1 answer
1 answer