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?
 A: $\begin{split}\lim\limits_{x\to0^-}\frac{e^{40x}-1}{5x} & = \lim\limits_{x\to0^-}\frac{(1+\frac{40x}{1!}+\frac{(40x)^2}{2!}+\cdots)-1}{5x} \\ &=\lim\limits_{x\to0^-}\frac{\frac{40x}{1!}+\frac{(40x)^2}{2!}+\cdots}{5x} \\ &=\lim\limits_{x\to0^-}\biggl(\frac{8}{1!}+\frac{40^2x}{2!\cdot5}+\frac{40^3x^2}{3!\cdot5}+\cdots\biggr) \\&=8 \end{split}$
A: $$l=\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}$$
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:
$$l=\frac {40} 5=8$$
A: 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,
$$\lim_{x\to0}\frac{e^{40x}-1}{40x}=\lim_{z\to0}\frac{e^z-1}z=\left.e^z\right|_{z=0}=1.$$
A: $$\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}\overset{\text{L'Hospital}}{=}\lim_{x\to 0^-} \frac{40e^{40x}}{5}=\frac{40\times1}{5}=8$$
A: Here is a short computation:
$$\lim_{x\to 0^{-}}\frac{e^{40x}-1}{5x}=\frac15\lim_{x\to 0^{-}}\frac{e^{40x}-1}{x}=\frac15\Bigl(\mathrm e^{40x}\Bigr)'_{x=0}=\frac{40\,\mathrm e^0}5=8.$$
A: Or back to basics. Let $f(x)=e^{40x}$. Then
$$
40=f'(0)=\lim_{x\to 0}\frac{e^{40x}-1}{x-0}=5 \lim_{x\to 0}\frac{e^{40x}-1}{5x}
$$
Therefore,
$$
\lim_{x\to 0}\frac{e^{40x}-1}{5x} = 8
$$
