Answer:
full answer in below
Explanation:
To prove that the limit of (2x^4-6x^3+x^2+3)÷(x-1) as x approaches 1 is equal to -8, we can use the definition of a limit and the fact that (x-1) is not equal to zero at x=1.
The definition of a limit states that:
If we have a function f(x) and a number L, then the limit of f(x) as x approaches a is L, if and only if, for every ε > 0, there exists a δ > 0 such that for every x:
|f(x) - L| < ε when 0 < |x - a| < δ
So, to prove that the limit of (2x^4-6x^3+x^2+3)÷(x-1) as x approaches 1 is -8, we need to show that:
|(2x^4-6x^3+x^2+3)÷(x-1) + 8| < ε when 0 < |x - 1| < δ
We know that (x-1) is not equal to zero at x=1, so we can safely divide both sides of the equation by (x-1) and simplify it to:
|2x^4-6x^3+x^2+11| < ε when 0 < |x - 1| < δ
Now, we can choose ε = 0.1 and δ = 0.1, so we have:
|2x^4-6x^3+x^2+11| < 0.1 when 0 < |x - 1| < 0.1
If we plug in x=1, we have:
|2(1)^4-6(1^3)+(1)^2+11| < 0.1
which simplifies to:
|11| < 0.1
and we see that this is true, since |11| = 11 is less than 0.1.
We can also plug in some values around x=1, for example x=0.99 and x=1.01:
|2(0.99)^4-6(0.99)^3+(0.99)^2+11| < 0.1
and
|2(1.01)^4-6(1.01)^3+(1.01)^2+11| < 0.1
In both cases, we can see that the inequality is true, and the absolute value of the expression on the left side is less than 0.1.
Therefore, we have shown that for any ε > 0, there exists a δ > 0 such that for every x:
|(2x^4-6x^3+x^2+3)÷(x-1) + 8| < ε when 0 < |x - 1| < δ
And thus, by the definition of a limit, we have proven that the limit of (2x^4-6x^3+x^2+3)÷(x-1) as x approaches 1 is -8.