HELP....last question SOS HELP

Consider the statement lim X-->1

X©ø-6X©÷+11X-6 / x-1 = 2

USING THE DEFINITIONS 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
- Next verify the limit is true by finding ©£ as an expression of ¥å
- Draw a picture illustrating the relation between ¥å, ©£ and the function
please leowdicap at gmail dot com

To understand what must be true for the given limit to hold, we need to consider the definitions of the limit.

Definition of a Limit: If f(x) approaches a unique number L as x approaches a from both sides, we write:

lim(x->a) f(x) = L

Now let's analyze the given expression and see what conditions must be met for the limit to hold:

lim(x->1) (x^3-6x^2+11x-6) / (x-1) = 2

For the limit to hold, the numerator and the denominator must approach their respective values as x approaches 1. In this case, the denominator (x-1) must approach 0 because it cannot be divided by zero.

So, we have one condition for the limit to hold:

x - 1 ≠ 0

This implies x ≠ 1. Therefore, the value of x should not be equal to 1.

To verify the limit is true by finding an expression of ϵ as a function of δ, we need to rewrite the given expression and simplify it:

(x^3-6x^2+11x-6) / (x-1) = [(x-1)(x^2-5x+6)] / (x-1)

Notice that (x-1) appears both in the numerator and denominator, which allows us to cancel it out:

(x^2-5x+6)

Now, we can simplify further:

(x-3)(x-2)

So, the limit expression becomes:

lim(x->1) (x-3)(x-2)

To evaluate this limit, we can substitute x = 1 into the expression:

(1-3)(1-2) = (-2)(-1) = 2

We find that the limit evaluates to 2, which matches the original statement.

To draw the relation between ϵ (epsilon), δ (delta), and the function, we first define the distance from the limit L, which is 2 in this case, to the function f(x). Let's call this distance ϵ.

We can see that as x gets closer to 1, the function (x-3)(x-2) approaches 2. This means that for any positive ϵ (epsilon), there exists a positive δ (delta) such that if the distance between x and 1 is smaller than δ, the distance between f(x) and 2 (ϵ) will also be smaller than ϵ.

In graphical terms, we can plot the function (x-3)(x-2) and draw a horizontal line at y = 2. The line represents the value of the limit L. As x approaches 1, the function gets closer and closer to the line at y = 2, indicating that the limit holds.

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