Why is $dL = L d\epsilon$?

Let's say there's a random elastic material. It's length is $L$ and it's tensile strain $\epsilon= (L-L_0)/L_0$ Now, when one pulls on it the following is true: $dW_{tot}=FdL =\sigma AdL=\sigma A L d\epsilon$

Why can we make that last step or in other words: why is $dL = L d\epsilon$?

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From the above definition for tensile strain, $L$ is $L=\epsilon L_{0}+L_{0}$. Thus

$dL=L_{0}d\epsilon$

replacing in the above $L_{0}$ with $L_{0}=\frac{L}{1+\epsilon}$ you get

$dL=\frac{L}{1+\epsilon}d\epsilon$

Now, if the deformation of the material is very small you can expand $\frac{1}{1+\epsilon}$ as

$\frac{1}{1+\epsilon}=1-O(\epsilon)$

Hence you find that $dL=Ld\epsilon$

EDIT:

The last step is possible because the series for $\frac{1}{1+\epsilon}$ for $|\epsilon|<1$ is $1-\epsilon +\epsilon^{2}-\epsilon^{3}+\dots$. From $|\epsilon|<1$ you find the condition for $L/L_{0}$ in which you can use this expansion.

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Can you please elaborate a bit more on the last step. Why can you expand it only if the deformation is very small? – Edward Stumperd Mar 19 '13 at 4:38
Nijankowski: The $O(\epsilon)$ should not be squared. @Edward: it only makes sense to truncate the series after the first term if the subsequent terms are small since $1/(1+\epsilon) = 1-\epsilon+O(\epsilon^2)$. – Ramashalanka Mar 19 '13 at 4:39
@Ramashalanka my bad, was thinking of something else, thanks for pointing it out. – nijankowski Mar 19 '13 at 4:58
@EdwardStumperd i've edited my answer. Hope its clear now. – nijankowski Mar 19 '13 at 4:58