Is this sequence of functions equicontinuous?

Question

$x_a = e^{t-a}, t \in [0,1], a \in \mathbb{R}, a \geq 0$ I've shown $x_a$ is uniformly bounded by e, now I just need to show if its equicontinuous, but I'm having trouble. I think its not.

Let $\delta >0$ be given and $t,y \in [0,1]$. Take $t=0$ and $y = \frac{1}{N} \in (0,\delta)$ where N is sufficiently large. Then $|e^{-a} - e^{y-a}| = |e^{-a}||1-e^y|$, but I can't figure it out from here

Answer 1

Let $t_0\in [0,1]$ be fixed and choose some $\epsilon > 0$.

Notice that $$|e^{t-a} - e^{t_0-a}| = e^{-a}|e^t - e^{t_0}| \leq |e^t-e^{t_0}|$$ for every $a\geq 0$, as $0\leq e^{-a}\leq 1$ for every $a\geq 0$.

Since $t\mapsto e^t$ is continuous, there exists $\delta >0$ such that $$|t-t_0|<\delta\quad \implies \quad|e^t-e^{t_0}|<\epsilon$$

Such $\delta$ do the job for equicontinuity.

Answer 2

What is the derivative of $x_a$?. The derivative is $x_a$ itself right? The fact that $\{x_a :a \geq0 \}$ is uniformly bounded implies the derivatives are also bounded and Mean value Theorem tells you that the family is equi-continuous.