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So I am supposed to find the limit

$\lim_{x\rightarrow\infty}\dfrac{\ln(1+x+x^\frac{1}{3})}{\ln(1+x^\frac{1}{4}+x^\frac{1}{3})}$,

however I can't seem to simplify it to something I know how to solve. I've tried various $ln()$ identities and tried to substitute for $x$ but cannot seem to get anywhere.

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    $\begingroup$ Did you try L'Hospital's Rule? That seems like it could be promising. $\endgroup$ Dec 24 '17 at 23:45
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    $\begingroup$ The logarithm is smooth, and so are roots like these. It's a quotient of a log of a sum of a continuous functions. You can check all those things are themselves differentiable, and so you're good :) $\endgroup$ Dec 24 '17 at 23:53
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Put $a = x^{\frac{1}{12}}$. Then limit becomes $$\lim_{a\to\infty}\dfrac{\ln(1+a^{12}+a^4)}{\ln(1+a^3+a^4)} = \lim_{a\to\infty}\bigg(\frac{12a^{11}+4a^3}{1+a^{12}+a^4} \cdot \frac{1+a^3+a^4}{3a^2+4a^3}\bigg) = \frac{12}{4} = 3$$ by L'Hospital's Rule (Notice that largest exponent in this expression is $15$ for both nominator and denominator and coefficients of them are $12$ and $4$ respectively).

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Hint:

Use first the substitution $t=x^{1/12}$. You can rewrite the expression as $$\frac{\ln(1+t^4+t^{12})}{\ln(1+t^{3}+t^4)},$$ then use equivalents.

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If you consider $$A=\frac{\ln(1+x+x^\frac{1}{3})}{\ln(1+x^\frac{1}{4}+x^\frac{1}{3})}$$ for infinitely large values of $x$, the dominant term is $x$ in numerator and $x^{1/3}$ in denominator. Then $$A=\frac{\ln(1+x+x^\frac{1}{3})}{\ln(1+x^\frac{1}{4}+x^\frac{1}{3})}\sim \frac{\ln(x)}{\ln(x^\frac{1}{3})}=\frac{\ln(x)}{\frac 13\ln(x)}=3$$

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One can use the substitution $x=1/t^{12}$ so that $t\to 0^{+}$. The expression under limit is transformed into $$\frac{\log(1+t^8+t^{12})-12\log t} {\log (1+t+t^4)-4\log t} $$ which can be further written as $$\dfrac{\dfrac{\log (1+t^8+t^{12})}{\log t} - 12}{\dfrac{\log(1+t+t^{4})}{\log t} - 4}$$ Since $\log t\to-\infty$ it follows that the desired limit is $-12/(-4)=3$.

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$${\ln(1+x+x^{1/3})\over\ln(1+x^{1/4}+x^{1/3})}={\ln x+\ln(x^{-1}+1+x^{-2/3})\over{1\over3}\ln x+\ln(x^{-1/3}+x^{-1/12}+1)}=3\left({1+\displaystyle{\ln(x^{-1}+1+x^{-2/3})\over\ln x}\over1+\displaystyle{3\ln(x^{-1/3}+x^{-1/12}+1)\over\ln x}}\right)\to3\left(1+{\ln(0+1+0)\over\ln\infty}\over1+{3\ln(0+0+1)\over\ln\infty} \right)=3\left(1+{0\over\infty}\over1+{0\over\infty}\right)=3$$

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  • $\begingroup$ What does $\ln (\infty)$ mean? $\endgroup$
    – user223391
    Dec 25 '17 at 5:47

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