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|

only if c=3.

To prove the given limit statement, we need to show that for any positive epsilon (ε), there exists a positive delta (δ) such that if 0 < |x - 3| < δ, then |√(x) - √(c)| < ε.

Let's start by using the hint provided:

0 < |√(x) - √(c)| = |x - c| / (√(x) + √(c)) < (1 / √(c)) * |x - c|

We want to find a suitable value for δ in terms of ε. Notice that if we restrict |x - 3| < δ, then we also have |x - c| < |x - 3| + |3 - c|. Therefore, we can choose δ = min(1, ε * √(c)).

Now, let's substitute this value of δ into our inequality:

|x - c| < δ = min(1, ε * √(c))

We can further simplify this inequality by noticing that |x - c| = |x - 3 + 3 - c| ≤ |x - 3| + |3 - c|. Therefore, we have:

|x - c| < |x - 3| + |3 - c|

Using the triangle inequality, we can rewrite this inequality as:

|x - c| < |x - 3| + |3 - c| < δ + |3 - c|

Substituting the value of δ = min(1, ε * √(c)), we have:

|x - c| < min(1, ε * √(c)) + |3 - c|

Now, we can substitute this inequality into our hint:

0 < |√(x) - √(c)| = |x - c| / (√(x) + √(c)) < (1 / √(c)) * |x - c|

We can further simplify this inequality by multiplying both sides by (√(x) + √(c)), which is positive:

0 < |x - c| < (1 / √(c)) * |x - c| * (√(x) + √(c))

Now, since (√(x) + √(c)) is always positive, we can divide both sides of the inequality by (√(x) + √(c)):

0 < |x - c| / (√(x) + √(c)) < (1 / √(c)) * |x - c|

Notice that we have arrived back at the given hint inequality. Therefore, we can conclude that if we choose δ = min(1, ε * √(c)), then for any positive epsilon (ε), if 0 < |x - 3| < δ, we have |√(x) - √(c)| < ε.

Hence, we have established that lim x->3 (√(x)) = (√(c)).