$\lim_{x\to 0}\frac{x}{x+x^2e^{i\frac{1}{x}}} = 1$

Question

$$\lim_{x\to 0}\frac{x}{x+x^2e^{i\frac{1}{x}}} = 1$$

I've trying seeing that:

$$e^{\frac{1}{x^2}i} = 1+\frac{i}{x^2}-\frac{1}{x^42!}-i\frac{1}{x^63!}+\frac{1}{x^84!}+\cdots\implies$$

$$x^2e^{\frac{1}{x^2}i} = x^2+\frac{i}{1}-\frac{1}{x^22!}-i\frac{1}{x^43!}+\frac{1}{x^64!}+\cdots$$

but I couldn't find a way to prove that the limit goes to $1$

Answer 1

First some simplification and then assuming $\;x\in\Bbb R\;$:

$$\frac x{x+x^2e^{i/x}}=\frac1{1+xe^{i/x}}=\frac1{1+x\left(\cos\frac1x+i\sin\frac1x\right)}\xrightarrow[x\to0]{}\frac1{1+0}=1$$

since

$$\left|e^{i/x}\right|=\left|\cos\frac1x+i\sin\frac1x\right|\le1\;\;\;\text{and this is then a bounded expression}$$

Answer 2

We assume that $x \in \mathbb{R}$. One may write, as $x \to 0$, $$ \left|\frac{x}{x+x^2e^{i\frac{1}{x}}} \right|=\frac{1}{\left|1+xe^{i\frac{1}{x}} \right|}\le \frac{1}{1-\left|xe^{i\frac{1}{x}} \right|}=\frac{1}{1-\left|x\right|} \to 1 $$ since $$ ||a|-|b||\le |a+b|. $$