4x^2 / 2x^2?

There is a mathematical property that says:
a^m / a^n = a^(m-n)

In this case, "a" is x. And you can divide 4/2 = 2 in front of this. So you result in 2 * x ^ (2-2) = 2 * x^0 = 2

Your answer is right

It is the x² cancel out and leaves you with 4/2 which is 2

you subtract the exponents when you're dividing.
So 2-2=0
Therfore the answer is 2 :)

4. It is because I am really smart.

4 x²
------
2 x²

2

4x^2/2x^2
= (4/2)(x^2/x^2)
= 2(1)
= 2

It is in fact 2.

x^2 / x^2 will always be 1. So we can assume that it is 1 / 1. Now you can add the constants back in. We take 4 * 1 / 2 * 1 = 4 / 2 = 2.

break it down into factors, then cancel out what can be canceled out.

(4)(x^2) / (2)(x^2) =

(4/2) (x^2/x^2) = (2)(1)

---


x^2 / x^2 =1
even when x = 0

one uses the limit test.

We know that if x is not zero, then x^2 / x^2 = 1

Limit (as x approaches zero) of x^2 / x^2 = 1

You can check with series that converge to zero:

for example:

Let x = 1/(2^n) for all n = 1 to infinity
Then Lim (as n grows to infinity) of 1/(4^2n) = 0

x^2 / x^2 = (4^2n)/(4^2n) = 1 for all values of n

Therefore
Limit (as x approaches 0) of x^2/x^2 = Lim (as n grows to infinity) of [1/(4^2n)] / [1/(4^2n)] = Lim (n to inf.) of 4^(2n) / 4^(2n) = 1

yes it's 2

Yes it is, only if x is not equal to 0.