# How can elasticity of $f(x)$ be expressed as a percentage change in $f(x)$ per $1$% change in $x$?

In answering the above question, we took the formula of elasticity:

$$El_xf(x) = (x/f(x))*(∂f(x)/∂x)$$

We can transform it in the following way:

$$El_xf(x) = (∂f(x)/f(x))*(∂x/x)$$

Now at this point, the professor took the limit of f(x) in respect to f(x). I can't seem to understand how could he come up with the following limit: $$lim_{Δ->0} (f(x+Δ)-f(x))/ f(x) / lim_{Δ->0} ((x+Δ)-x)/ x$$

It seemed he took the derivative (limit version) in respect to f(x). I would appreciate any help!

Thank you!

The (average) elasticity of $$f(x)$$ w.r.t. $$x$$ (in the interval $$[x,\, x+\Delta x$$]) is defined by \begin{align*} E&=\frac{\text{percentual change in }f(x)}{\text{percentual change in }x}\\ &=\frac{\Delta f(x)/f(x)}{\Delta x/x}\\ &=\frac{x}{f(x)}\cdot\frac{\Delta f(x)}{\Delta x}\\ &=\frac{x}{f(x)}\cdot\frac{f(x+\Delta x)- f(x)}{\Delta x} \end{align*} For example, if $$\frac{\Delta x}{x}=1\%$$, and $$E=3$$, then the percentual change in $$f$$ is $$\frac{\Delta f}{f}=3\cdot1\%=3\%$$.
For $$\Delta x \to 0$$, $$E=\frac{x}{f(x)}\lim_{\Delta x\to 0}\frac{f(x+\Delta x)- f(x)}{\Delta x}=\frac{x}{f(x)}f'(x)=\frac{\mathrm d \log f(x)}{\mathrm d \log x}$$ and is called point elasticity (generally the symbol is $$\varepsilon$$ or $$\eta$$).