lim-x-4-x-2-3-x-2-x-4-x-1-3-x-1-

Question Number 144106 by mathdanisur last updated on 21/Jun/21

\underset{x \to \infty}{l i m} \frac{4^{x - 2} + 3^{x} + 2^{x}}{4^{x - 1} + 3^{x + 1}} = ?

Commented by bobhans last updated on 22/Jun/21

\lim_{x \to \infty} \frac{4^{x} (4^{- 2} + {(\frac{3}{4})}^{x} + {(\frac{1}{2})}^{x})}{4^{x} (4^{- 1} + 3 {(\frac{3}{4})}^{x})}

= \lim_{x \to \infty} (\frac{4^{- 2} + {(\frac{3}{4})}^{x} + {(\frac{1}{2})}^{x}}{4^{- 1} + 3 {(\frac{3}{4})}^{x}})

= \frac{4^{- 2}}{4^{- 1}} = 4^{- 1} .

Answered by Dwaipayan Shikari last updated on 21/Jun/21

\lim_{x \to \infty} \frac{4^{x - 2} + 3^{x} + 2^{x}}{4^{x - 1} + 3^{x + 1}}

= \frac{\frac{1}{4^{2}} + {(\frac{3}{4})}^{x} + {(\frac{2}{4})}^{x}}{\frac{1}{4} + 3 {(\frac{3}{4})}^{x}} = \frac{\frac{1}{4^{2}} + 0 + 0}{\frac{1}{4} + 3 (0)} = \frac{1}{4}

Commented by harckinwunmy last updated on 21/Jun/21

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Commented by mathdanisur last updated on 21/Jun/21

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