Question

$\underset{x\to \infty }{\lim }{\left(\frac{3x-4}{3x+2}\right)}^{\frac{x+1}{3}}$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(\frac{3x-4}{3x+2}\right)}^{\left(\frac{x+1}{3}\right)} \\ =\underset{x\to \infty }{\lim }{\left(1+\frac{3x-4}{3x+2}-1\right)}^{\left(\frac{x+1}{3}\right)} \\ =\underset{x\to \infty }{\lim }{\left(1+\frac{3x-4-3x-2}{3x+2}\right)}^{\left(\frac{x+1}{3}\right)} \\ =\underset{x\to \infty }{\lim }{\left(1-\frac{6}{3x+2}\right)}^{\left(\frac{x+1}{3}\right)} \\ \mathrm{Using}\mathrm{the}\mathrm{theorem}\mathrm{given}\mathrm{below}: \\ \mathrm{If}\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)=0\mathrm{such}\mathrm{that}\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\mathrm{exists},\mathrm{then}\underset{x\to a}{\lim }{\left[1+f\left(x\right)\right]}^{\frac{1}{g\left(x\right)}}=e^{\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}}. \\ \mathrm{Here}: \\ f\left(x\right)=-\frac{6}{3x+2} \\ g\left(x\right)=2x \\ \Rightarrow e^{\underset{\mathrm{x}\to \infty }{\lim }\left(\frac{-6}{3\mathrm{x}+2}\times \frac{\mathrm{x}+1}{3}\right)} \\ =e^{\underset{\mathrm{x}\to \infty }{\lim }\left(\frac{-6\mathrm{x}-6}{9\mathrm{x}+6}\right)} \\ =e^{\underset{\mathrm{x}\to \infty }{\lim }\left(\frac{-6-{\displaystyle \frac{6}{x}}}{9+{\displaystyle \frac{6}{x}}}\right)} \\ =e^{-\frac{2}{3}}(\mathrm{When}x\to \infty ,\frac{1}{\mathrm{x}}\to 0) \end{gathered}$$