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$\lim_{x\rightarrow \infty }\left(4^{x}+5^{x}\right)^{1/x}$
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3 Answers

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$\lim_{x \to \infty} (4^x+5^x)^\frac{1}{x} = \lim_{x \to \infty} \left(5^x \left((\frac{4}{5})^x +1\right )\right)^\frac{1}{x}\\ =\lim_{x \to \infty} 5\times( )^{\frac{1}{\infty}} \text {(the term inside () ranges between 1 and 2)} \\ =5\times()^0\\ =5\times1\\=5$
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very easy

lt->inf (4^x+5^x)^1/x

lt->inf 5^x*1/x[(4/5)^x +1]^1/x

ans will be 5

since 4/5 will give <1 no. and its to the power inf will results 0
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$\bf{\displaystyle \lim_{x\rightarrow \infty} \left(4^x+5^x \right) \approx \lim_{x\rightarrow \infty}5^x}$

$\bf{So\; \displaystyle  \lim_{x\rightarrow \infty} \left(4^x+5^x \right)^{\frac{1}{x}} \approx \lim_{x\rightarrow}\left(5\right)^{\frac{x}{x}} = 5.}$

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