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$$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=?$$

I have tried a lot but I am stuck when I solve this by using this hints. $a^x=\exp(\ln(a^x))=\exp(x\ln a)$, so then $a=\frac{4^x+5^x}{4}$. The above expression becomes $$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=\lim_{x\to\infty}\exp\left(\ln\left(\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}\right)\right)=\exp\left(\lim_{x\to\infty}\frac{1}{x}\ln\left(\frac{4^x+5^x}{4}\right)\right).$$ Now here I am stuck that how I can deal this indeterminate expression?

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2 Answers 2

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Do you have to use this hint?

You could simplify your task if you simplify the given expression first: $$\frac{4^x+5^x}{4}=\frac{1}{4}\cdot(4^x+5^x)=\frac{1}{4}\cdot5^x\cdot\left(\left(\frac{4}{5}\right)^x+1\right),$$ therefore $$\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=\left(\frac{1}{4}\cdot5^x\cdot\left(\left(\frac{4}{5}\right)^x+1\right)\right)^{\frac{1}{x}}=\frac{1}{4^{\frac{1}{x}}}\cdot5\cdot\left(\left(\frac{4}{5}\right)^x+1\right)^{\frac{1}{x}}.$$

From this, the limit should be pretty evident.

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Let's assume that the required limit to be calculated is $L$ And by taking the natural logarithm both sides, we would have: $$\ln L =\lim_{ x\to \infty }\frac{\ln(\frac{4^x+5^x}{4})}{x}$$ This is an inderminate form of $\frac{\infty}{\infty}$ and can be solved by applying L' Hopital's Rule.
Applying the rule we get:
$$\ln L = \lim_{x\to\infty}\frac{\ln 5\cdot 5^x+\ln 4\cdot 4^x }{5^x+4^x}$$ Dividing the numerator and the denominator by $5^x$ we have: $$\ln L = \lim_{x\to \infty} \frac{\ln 5+ \ln 4\cdot (\frac45)^x}{1+(\frac{4}{5})^x}$$ Since a number less than one when raises to power tending to infinity tends to $0$ , we have : $$\ln L = \ln 5$$ Which means : $$L =5$$

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