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

$$\underset{x\to \mathrm{\infty }}{lim}\frac{4^{x-2}+3^x+2^x}{4^{x-1}+3^{x+1}}=?$$

Commented by bobhans last updated on 22/Jun/21

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

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

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{4^{x-2}+3^x+2^x}{4^{x-1}+3^{x+1}}$$$$=\frac{\frac{1}{4^2}+{\left(\frac{3}{4}\right)}^x+{\left(\frac{2}{4}\right)}^x}{\frac{1}{4}+3{\left(\frac{3}{4}\right)}^x}=\frac{\frac{1}{4^2}+0+0}{\frac{1}{4}+3\left(0\right)}=\frac{1}{4}$$

Commented by harckinwunmy last updated on 21/Jun/21

$$ff$$

Commented by mathdanisur last updated on 21/Jun/21

$$thanksSir$$

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