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For small $\epsilon \geq 0$ Is

$erf(\epsilon) \leq \epsilon$

Can somebody give me the hint

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Hint: One definition of the error function is $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}\,dt.$$

We always have $e^{-t^2}\le 1$, with equality only at $0$, so for positive $\epsilon$, $$\operatorname{erf}(\epsilon)\lt \frac{2}{\sqrt{\pi}}\epsilon.$$ However, $\dfrac{2}{\sqrt{\pi}}\gt 1$.

And for $t$ close to $0$, $e^{-t^2}$ is very close to $1$. For example, calculate $e^{-t^2}$ for $t=0.01$.

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What is the derivative of the error function in the vicinity of zero? And what is the sign of the second derivative?

More generally, what does the Taylor series look like at zero?

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