First slide

Introduction to limits

Question

The value of $lim_{x \to \infty} {(\frac{3 x - 4}{3 x + 2})}^{(\frac{x + 1}{3})}$ ,is

Moderate

A

$e^{- 2 / 3}$
B

$e^{- 1 / 3}$
C

$e^{- 2}$
D

$e^{- 1}$

Solution

We have,
$\begin{array}{l} lim_{x \to \infty} {(\frac{3 x - 4}{3 x + 2})}^{\frac{x + 1}{3}} = lim_{x \to \infty} {(1 + \frac{- 6}{3 x + 2})}^{\frac{x + 1}{3}} \\ = e^{lim_{x \to \infty} \frac{- 6}{3 x + 2} \times \frac{x + 1}{3}} = e^{lim_{x \to \infty} \frac{- 2 x - 2}{3 x + 2}} = e^{- 2 / 3} \end{array}$

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