# How do you prove limit of #root3(x)# as #x->0# using the precise definition of a limit?

For any

Considering that:

so:

which proves that:

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To prove the limit of √3(x) as x approaches 0 using the precise definition of a limit, we need to show that for any positive ε, there exists a positive δ such that if 0 < |x - 0| < δ, then |√3(x) - L| < ε, where L is the limit we are trying to prove.

Let's start by assuming that |√3(x) - L| < ε. We can square both sides of this inequality to get rid of the square root:

(√3(x) - L)^2 < ε^2

Expanding the left side:

3x - 2√3(x)L + L^2 < ε^2

Now, we can focus on finding a suitable δ. Notice that 3x - 2√3(x)L + L^2 is a quadratic expression in terms of x. We want to find a δ such that if 0 < |x - 0| < δ, then the above inequality holds.

To simplify the expression, we can make an assumption that |x| < 1, which implies -1 < x < 1. This allows us to approximate √3(x) as 0 since x is approaching 0. With this approximation, the inequality becomes:

L^2 < ε^2

Taking the square root of both sides:

|L| < ε

This tells us that if we choose δ = 1, then for any 0 < |x - 0| < δ, we have |√3(x) - L| < ε. Therefore, we have proven that the limit of √3(x) as x approaches 0 is L, where L can be any value satisfying |L| < ε.

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