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?

Your mistake is to consider that
$$\lim_{x\to0}\frac{e^{40x}-1}{40x}$$ is the derivative of $e^{40x}$ on $x$. But the derivative of $e^{40x}$ on $x$ would be
$$\lim_{x\to0}\frac{e^{40x}-1}{x}.$$
In other words,
$$\lim_{x\to0}\frac{e^{40x}-1}{40x}=\lim_{z\to0}\frac{e^z-1}z=\left.e^z\right|_{z=0}=1.$$
