I am trying to show that the approximation on $0\leq x \leq 1$ $$\phi(x,\epsilon) \sim \sin x+ \epsilon \cos x - \epsilon$$ is uniformly asymptotic to the exact solution $$f(x,\epsilon) = \frac{1}{1+\epsilon}\left[\sin x +\epsilon \cos x - \epsilon e^{-\epsilon x}\right]$$. The definition of uniformly asymptotic: Suppose $f(x,\epsilon)$ and $\phi(x,\epsilon)$ are continuous functions for $x \in I$, $0 < \epsilon < \epsilon_0$. Then $\phi$ is a uniformly valid asymptotic approximation of $f$ if for all $x \in I$ and $\delta > 0$, there is $\epsilon_1$ such that $|f(x,\epsilon)-\phi(x,\epsilon)|\leq\delta |\phi(x,\epsilon)|$, $0< \epsilon< \epsilon_1$
So here is what I have:
I subtract the two solutions and get $$\left|\frac{-\epsilon}{1+\epsilon} \sin x + \frac{-\epsilon^2}{1+\epsilon} \cos x - \epsilon(e^{-\epsilon x} -1)\right| \leq \delta \left|(\sin x+ \epsilon \cos x - \epsilon)\right|$$ Now if I take the limit as $\epsilon \to 0$ I get that $$|0 + 0 - 0| \leq \delta |\sin x+ 0 - 0| \Rightarrow 0 \leq \delta |\sin x|$$ But for $0\leq x \leq 1$, $|\sin x| > 0$ and $\delta > 0$ so the inequality holds. Is this sufficient to prove that the approximation is uniformly asymptotic?