Showing an approximation is uniformly asymptotic 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? 
 A: HINT No, it's not: there's clearly something going on around $(x,\varepsilon)=(0,0)$ which needs to be resolved, plus a subtle point relating to the zeroes of your RHS. A plan of attack would read like this:


*

*Define
$$
g(x,\varepsilon)
=
\frac{
(1+\varepsilon)^{-1} \sin(x)
+
\varepsilon (1+\varepsilon)^{-1} \cos(x)
-
(1-{\rm e}^{-\varepsilon x})
}{
\sin(x) +
\varepsilon \cos(x) - \varepsilon
} .
$$
Your inequality reads $\sup_{(x,\varepsilon)\in[0,1]\times(0,\varepsilon_1)} \left\vert\varepsilon \, g(x,\varepsilon) \right\vert < \delta$, where $\delta>0$ is given and  $\varepsilon_1$ can be taken as small as needed.

*Plot the above two-variable function to gain some intuition.

*Focus on the point $(x,\varepsilon)=(0,0)$, which appears problematic: it makes the denominator zero. You thus have to determine whether $\lim_{(x,\varepsilon)\to(0,0)} \left\vert\varepsilon \, g(x,\varepsilon) \right\vert$ exists. Even worse, the denominator may be expected to have an entire curve of zeroes; it's "improbable" (chuckle...) that the numerator will be zero on that precise same set, and you can't bound something "small" by something zero :).
