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  4. The value of displaystyle limx → ∞ ((3x-4/3x+2))(x+1/3) is equal to :

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Q. The value of $\displaystyle \lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} $ is equal to :

Limits and Derivatives

Solution:

Consider $\displaystyle \lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} = \displaystyle \lim_{x \to \infty }\left(\frac{3x+2-6}{3x+2}\right)^{\frac{x+1}{3}}$
$= \displaystyle \lim_{x \to \infty }\left(\frac{3x+2}{3x+2}-\frac{6}{3x+2}\right)^{\frac{x+1}{3}} = \displaystyle \lim_{x \to \infty }\left(1+\frac{-6}{3x+2}\right)^{\frac{x+1}{3}}$
$= \displaystyle \lim_{x \to \infty }\left(\left\{1+\frac{-6}{3x+2}\right\}^{\frac{3x+2}{-6}}\right)^{\frac{-6}{3x+2}\times\frac{x+1}{3}}$
$= \displaystyle \lim_{x \to \infty }\left[e\right]^{\frac{-6}{3x+2}\times \frac{x+1}{3}} \left(\because \displaystyle \lim_{x \to \infty }\left(1+\frac{1}{x}\right)^{x} = e\right)$
$= \displaystyle \lim_{x \to \infty } e^{\frac{-2x-2}{3x+2}} = \displaystyle \lim_{x \to \infty }e^{\frac{x\left(-2-2/x\right)}{x\left(3+2/x\right)}} = \displaystyle \lim_{x \to \infty }e^{\frac{-2-2/x}{3+2/x}} = e^{-2/3}$