Uniform Continuity of $\frac{e^{-\frac{1}{x}}}{x}$

Question

Is $\frac{e^{-\frac{1}{x}}}{x}$ uniformly continuous in the interval $(0,1)$? I'm having trouble evaluating the limit at $x\to 0$, hence I doubt its continuity to the right of $0$. Any ideas. Thanks.

Answer 1

To treat the limit as $x\to 0$ it suffices to use the mother of all inequalities about the exponential: $$\tag1e^t\ge 1+t\qquad\text{for all }t\in\Bbb R$$ and of course $e^{t+u}=e^te^u$. This makes $$e^t=(e^{t/2})^2\ge(1+\tfrac t2)^2=1+t+\frac14t^2\qquad\text{for all }t\ge-1$$ and so $$0<\frac{e^{-\frac1x}}x=\frac1{xe^{\frac1x}}\le \frac{1}{x+1+\frac1{4x}}\qquad\text{for all }x>0.$$ As $x\to 0^+$, the upper bound tends to $0$ as well.

Answer 2

Hint/Solution: lim$_{x \to 0+} \frac {e^{-1/x}}{x}=$lim $_{t \to \infty} te^{-t}=0$ (since exponential grows much faster than polynomial) and lim$_{x \to 1-} \frac {e^{-1/x}}{x}=e^{-1}$ exists.Hence...