Limit of an Exponential Equation

Question

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?

Answer 1

$$\lim_{x\to \infty} \bigg(1-\frac{4}{x}\bigg)^{3x} =\lim_{x\to \infty}e^{3x\ln(1-4/x)}$$

From here we only need to take the limit of the exponent, so $$\lim_{x\to \infty} 3x\ln(x-4/x) = 3\lim_{x\to \infty}\frac{\ln(1-4/x)}{1/x}$$ Here we can use l'hopitals rule since it is of the form $\frac{0}{0}$ to get

$$3\lim_{x\to \infty}\frac{4/x^{2}}{(-1/x^{2})(1-4/x)}=3\lim_{x\to \infty} \frac{-4}{1-4/x} =-4\cdot3= -12$$ and so we get $$\lim_{x\to \infty} \bigg(1-\frac{4}{x}\bigg)^{3x}=e^{-12}$$

Answer 2

Since $f(x)=x^{-12}$ is a continuous function in $x=e$, we obtain:

$$\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{3x}=\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}\cdot\left(-12\right)}=\left(\lim_{x \rightarrow \infty} \left(1-\frac{4}{x}\right)^{-\frac{x}{4}}\right)^{-12}=e^{-12}.$$