<s>[INST] Limit of an Exponential Equation My problem is the $\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}$. I got $$\ln(B) = \ln\left(\lim_{x \rightarrow \infty} 3x \cdot \ln\left(1-\frac{4}{x}\right)\right)$$
which equates to $$\frac{\ln\left(1-\frac{4}{x}\right)}{\frac{1}{3x}} = \frac{0}{0}$$
   so I can use L'Hôpital's rule
$$\frac{1}{1-\frac{4}{x}} \cdot \frac{1}{3x} - \ln\left(1-\frac{4}{x}\right) \cdot \left(-\frac{1}{3x^2}\right)$$
I'm stuck here. Am I even doing this correctly? If I am what are the next steps?
 [/INST] Since $f(x)=x^{-12}$ is a continuous function in $x=e$, we obtain:
$$\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}=\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}\cdot\left(-12\right)}=\left(\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}}\right)^{-12}=e^{-12}.$$
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