$$\lim_{n \to \infty}\left(\left(\frac{4}{3}\right)^{n} + 1\right) = \infty$$ $$\lim_{n \to 0^-}\left(\left(\frac{4}{3}\right)^{n} + 1\right) = 2$$ More at n→0 from the left $$\lim_{n \to 0^+}\left(\left(\frac{4}{3}\right)^{n} + 1\right) = 2$$ More at n→0 from the right $$\lim_{n \to 1^-}\left(\left(\frac{4}{3}\right)^{n} + 1\right) = \frac{7}{3}$$ More at n→1 from the left $$\lim_{n \to 1^+}\left(\left(\frac{4}{3}\right)^{n} + 1\right) = \frac{7}{3}$$ More at n→1 from the right $$\lim_{n \to -\infty}\left(\left(\frac{4}{3}\right)^{n} + 1\right) = 1$$ More at n→-oo