Limit $\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}$ $$\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}$$
What I have does is:
$$\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}=\lim_{x\to 0^{-}}\frac{e^{40x}-e^0}{5x}=\lim_{x\to 0^{-}}\frac{8}{8}\cdot\frac{e^{40x}-e^0}{5x}=\lim_{x\to 0^{-}}8\cdot\frac{e^{40x}-e^0}{40x-0}=\\=8\cdot(e^{40x})'_{x=0}=8\cdot40e^{40\cdot 0}=320$$
Which is incorrect, where is the problem?

$$l=\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}$$
Using the derivative definition:
$$l=\frac 1 5\lim_{x\to 0^{-}}\frac{e^{40x}-e^{40*0}}{x-0}$$
$$l=\frac 1 5(e^{40x})_{x=0}'$$
$$l=\frac 1 5(40e^{40x})_{x=0}$$
Therefore:
$$l=\frac {40} 5=8$$
