I need to solve a limit without using l'Hopital rule

Question

i need to solve this limit without using l'Hopital rule. Thanks in advance if you'll answer. $$\lim\limits _{x\to 0}\left(\frac{e^{-4x}-1}{x^2-x}\right)$$

Answer 1

Note that

$$\frac{e^{-4x}-1}{x^2-x}=\frac1{e^{4x}}\frac{1-e^{4x}}{4x}\frac{4x}{x^2-x}=\frac1{e^{4x}}\frac{e^{4x}-1}{4x}\frac{4}{1-x}\to1\cdot1 \cdot 4=4$$

Answer 2

$\lim\limits_{x\rightarrow 0}{{e^{-4x}-1}\over x}=-4$ (the derivative of $e^{-4x}$ at $0$), $\lim\limits_{x\rightarrow 0}{x\over {x^2-x}}=-1$. Take the product of the limits.

Answer 3

\begin{align*} \dfrac{e^{-4x}-1}{x^{2}-x}&=\dfrac{(1-4x+(16x^{2}/2)-\cdots)-1}{x^{2}-x}\\ &=\dfrac{8x-4-\cdots}{x-1}\\ &\rightarrow 4 \end{align*} as $x\rightarrow 0$.

Answer 4

That is my answer for the limit, it is a image...