Evaluate $\lim_{x\to 0}\frac{e^{-32x}-1}{4x}$ - Without L'Hospital

Question

Evaluate $$\lim_{x\to 0}\frac{e^{-32x}-1}{4x}$$

I have tried to use the squeeze theorem and taking $$\lim_{x\to 0}e^{\ln(\frac{e^{-32x}-1}{4x})}$$

Answer 1

It is ${1\over 4}f'(0)$ where $f(x)=e^{-32x}$,

$$f'(0)=\lim_{x\rightarrow 0}{{f(x)-f(0)}\over{x-0}}=\lim_{x\rightarrow 0}{{e^{-32x}-e^{(-32)0}}\over{x-0}}=\lim_{x\rightarrow 0}{{e^{-32x}-1}\over x}$$

Answer 2

$$\lim_{x\to 0}\frac{e^{-32x}-1}{4x}=\lim_{x\to 0}\frac{e^{-32x}-1}{-32x}\cdot\frac{-32x}{4x}=-8$$ I used the well-known limit $$\lim_{h\to 0}\frac{e^{h}-1}{h}=1$$

setting $h↦-32x$ gives the desired result