show lim x->3 (sqrt(x)) = (sqrt(c))
hint: 0< |sqrt(x)-sqrt(c)|= (|x-c|)/(sqrt(x)+sqrt(c))<(1/(sqrt(c))*|x-c|
lim x->c (sqrt(x)) = (sqrt(c))
means that the closer x is chosen to c the closer sqrt(x) comes to sqrt(c). To prove that, you need to show that making |sqrt(x) - sqrt(c)| smaller than some arbitrary epsilon is always possible by restricting x to some interval around c. Since epsilon can be any number larger than zero, this means that this interval around zero exists no matter how small epsilon is chosen. So, even if epsilon = 0.000000000001, there is stll a small interval around c for which the square root function is within 0.000000000001 of c.
From the inequalities:
0< |sqrt(x)-sqrt(c)|= (|x-c|)/(sqrt(x)+sqrt(c))<(1/(sqrt(c))*|x-c|
you can see that to require
|sqrt(x)-sqrt(c)| < epsilon
it is enough to require that:
|x-c| < sqrt(c) epsilon
So, we see that for arbitrary
epsilon > 0 there indeed does exist a delta > 0 such that for |x-c| < delta we have that
|sqrt(x) - sqrt(c)| < epsilon
To show that the limit of √x as x approaches 3 is equal to √c, we can use the epsilon-delta definition of a limit. According to this definition, for any ε > 0, there exists a δ > 0 such that if 0 < |x - 3| < δ, then |√x - √c| < ε.
Let's break down the hint provided:
0 < |√x - √c| = (|x - c|) / (√x + √c) < (1/√c) * |x - c|
The first inequality states that the absolute difference between √x and √c is equal to the absolute difference between x and c, divided by the sum of √x and √c.
The second inequality simplifies the denominator (√x + √c) to just √c, as √x will always be smaller than √c when x is approaching 3.
Now, let's proceed with the proof:
Given any ε > 0, we want to find a δ > 0 such that if 0 < |x - 3| < δ, then |√x - √c| < ε.
Starting with the hint:
0 < |√x - √c| = (|x - c|) / (√x + √c) < (1/√c) * |x - c|
We can see that if |x - c| < δ, then it implies that 0 < |x - 3| < δ.
Now, we want to find a value of δ such that (1/√c) * |x - c| < ε.
Let's consider the expression (1/√c) * |x - c|. If we set δ = ε * √c, we can see that:
(1/√c) * |x - c| < ε * √c
1 * |x - c| < ε * √c
|x - c| < ε * √c
Therefore, if 0 < |x - 3| < ε * √c (where δ = ε * √c), then it follows that |√x - √c| < ε.
Hence, we have shown that lim x → 3 (√x) = √c using the epsilon-delta definition of a limit.