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The value of  $\lim_{x\rightarrow \infty }\left ( \frac{4^{x+2} + 3^{x}}{4^{x-2}} \right )$ is
asked in Calculus by Active (1.3k points)
edited by | 105 views
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answer will  be  256

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Can you explain how did you got 256 ?
+2
256 + 16 ($\left ( \frac{3}{4} \right )^x$

now if x--> infinity  then $\left ( \frac{3}{4} \right )^x$ will go towards zero  hence only 256
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got it or not ??
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Understood. Thanks.

1 Answer

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$\left ( \frac{4^{x+2}+ 3^{x}}{4^{x-2}} \right )$

=$\left ( \frac{4^{x}4^{2} + 3^{x}}{\frac{4^{x}}{4^{2}}} \right )$

=$16 \left ( \frac{4^{x}*16+ 3^{x}}{4^{x}} \right )$

Taking $4^{x}$ common from numerator and denominator and then cancelling it.

=$16 \left ( \frac{16+ (\frac{3}{4})^{x}}{1} \right )$

$\therefore \textrm{value of  } _{x\rightarrow \infty }^{lim} \left ( \frac{4^{x+2}+ 3^{x}}{4^{x-2}} \right ) = _{x\rightarrow \infty }^{lim} 16 \left ( \frac{16+ (\frac{3}{4})^{x}}{1} \right ) = 16 \left ( \frac{16+0}{1} \right ) = 16*16 = 256$
answered by Loyal (8.3k points)

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