4. Prove the following using ϵ − δ definition: a. [2 marks] lim x→c x2 = c2. b. [4 marks] lim x→a ex = ea. 5?

a. To prove that lim x→c x^2 = c^2 using the ϵ − δ definition, we need to show that for any ϵ > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |x^2 - c^2| < ϵ.

Let ϵ > 0 be given. We want to find a δ > 0 such that if 0 < |x - c| < δ, then |x^2 - c^2| < ϵ.

|x^2 - c^2| = |x - c||x + c|. Since |x - c| < δ, we can choose δ = ϵ/(2|c| + 1). Then if 0 < |x - c| < δ, we have:

|x^2 - c^2| = |x - c||x + c| < δ|2c| = ϵ.

Therefore, lim x→c x^2 = c^2.

b. To prove that lim x→a e^x = e^a using the ϵ − δ definition, we need to show that for any ϵ > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |e^x - e^a| < ϵ.

Let ϵ > 0 be given. We want to find a δ > 0 such that if 0 < |x - a| < δ, then |e^x - e^a| < ϵ.

|e^x - e^a| = |e^a||e^(x-a) - 1|. Since |x - a| < δ, we can choose δ = ln(1 + ϵ)/2. Then if 0 < |x - a| < δ, we have:

|e^x - e^a| = |e^a||e^(x-a) - 1| < e^a|ln(1 + ϵ)/2| = ϵ.

Therefore, lim x→a e^x = e^a.

a. Let ϵ > 0 be given. We need to find a δ > 0 such that |x^2 - c^2| < ϵ whenever 0 < |x - c| < δ.

Since
$$|x^2 - c^2| = |(x + c)(x - c)| = |x + c||x - c|,$$

it suffices to show that |x + c| < 1 and |x - c| < 1. Since
$$|x + c| \le |x| + |c| \le |c| + |c| = 2|c|$$

and
$$|x - c| \le |x| + |c| \le |c| + |c| = 2|c|,$$

it suffices to take $\delta = \min\{1/2|c|, \epsilon/2\}$. Then, for 0 < |x - c| < δ, we have
$$|x + c| < 2|c| < 1 \quad \text{ and } \quad |x - c| < \epsilon/2.$$

Therefore,
$$|x^2 - c^2| = |(x + c)(x - c)| \le |x + c||x - c| < 1 \cdot \frac{\epsilon}{2} = \frac{\epsilon}{2} < \epsilon.$$

Thus, lim_(x->c) x^2 = c^2.

b. Let ϵ > 0 be given. We need to find a δ > 0 such that |e^x - e^a| < ϵ whenever 0 < |x - a| < δ.

Since e^x is continuous on R, it is uniformly continuous on [a - δ, a + δ] for any δ > 0. So, there exists a δ > 0 such that |e^x - e^a| < ϵ whenever |x - a| < δ.

Thus, lim_(x->a) e^x = e^a.