What would be the value of the limit $\lim _{x\to \infty} (\frac{3x +1}{3x-1})^{4x}$? What would be the value of the limit $\lim _{x\to \infty} (\frac{3x +1}{3x-1})^{4x}$?
My initial idea was to divide by x in the numerator and denominator. However that would only solve the inner problem. How can the power be manipulated to evaluate the limit?
 A: $$\left(\frac{3x +1}{3x-1}\right)^{4x}=\left[\frac{\left(1+\frac{1}{3x}\right)^{3x}}{\left(1-\frac{1}{3x}\right)^{3x}}\right]^\frac{4}{3}\to\left[\frac{e}{e^{-1}}\right]^{\frac{4}{3}}=e^\frac{8}{3}$$
A: HINT:
$$\dfrac{3x+1}{3x-1}=1+\dfrac2{3x-1}$$
Use $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$
$$\lim_{x\to\infty}\left(\frac{3x +1}{3x-1}\right)^{4x}=\left[\lim_{x\to\infty}\left(1+\dfrac2{3x-1}\right)^{\dfrac{3x-1}2}\right]^{\lim_{x\to\infty}\dfrac{2\cdot4x}{3x-1}}=?$$
A: You may transform the variable linearly ($3x-1=t$) to make a term disappear in the denominator and make the expression more familiar.
$$\lim_{x\to\infty}\left(\frac{3x+1}{3x-1}\right)^{4x}=\lim_{t\to\infty}\left(\frac{t+2}t\right)^{4(t+1)/3}=\lim_{t\to\infty}\left(1+\frac2t\right)^{4t/3+4/3}.$$
Then with a rescaling, you remove another coefficient
$$\lim_{t\to\infty}\left(1+\frac2t\right)^{4t/3}\left(1+\frac2t\right)^{4/3}=\lim_{u\to\infty}\left(1+\frac1u\right)^{8u/3}\cdot1=\left(\lim_{u\to\infty}\left(1+\frac1u\right)^u\right)^{8/3}.$$
