Wednesday

August 20, 2014

August 20, 2014

Posted by **Tommy feels dumb** on Wednesday, May 7, 2014 at 1:25pm.

lim x-6x+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 as an expression of ϵ.

Draw a picture illustrating the relation between ϵ, and the function.

i am at hulk422 at g mail . com

- Calculus (lim) -
**Steve**, Wednesday, May 7, 2014 at 5:02pmx^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.

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