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This is from Stewarts's Calculus Early Transcendentals 8e, Chapter 4, Problem Plus, #8. The topics discussed in chapter 4 are: "Maximum and Minimum Values", "The Mean Value Theorem", "How Derivatives Affect the Shape of a Graph", "Indeterminate Forms and L'Hospital's Rule", "Summary of Curve Sketching", "Graphing with Calculus and Calculators", "Optimization Problems", "Newton's Method", and "Antiderivatives".

The problem is: $$\lim_{x\to\infty}\frac{(x+2)^{1/x}-x^{1/x}}{(x+3)^{1/x}-x^{1/x}}$$

I found the limits of each term become 1, which makes the fraction $\frac{0}{0}$, so I tried using L'Hospital's but it doesn't really help as the derivatives become even more complicated. I tried rationalizing either top or bottom, but it's not too easy either because the exponents contain $x$. I also tried using Squeeze Theorem, but all failed. I think there is a way to rationalize this to simplify and then use L'Hospital's from there but I really can't find a way. I don't normally give up but as I've been struggling with this for a month, I think it's time to seek help from others :( According to graphing devices, it looks like the limit approaches $\frac{2}{3}$. I want to find a way to verify this only using elementary calculus (no series expansion, or etc) as I think that's how Stewart intended.

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  • $\begingroup$ Seeing $x$'s in exponents like that makes me want to start taking logarithms, in the spirit of the first answer to this question. That might help you get moving in the right direction. $\endgroup$ – Robert Howard Jul 3 '18 at 5:52
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Write the ratio as $\frac {(1+\frac 2 x )^{1/x} -1} {(1+\frac 3 x )^{1/x} -1}$. Thus you have to find $\lim_{y \to 0} \frac {(1+2y)^{y}-1} {(1+3y)^{y}-1}$. Can you take it from here?

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  • $\begingroup$ ah yes it worked! thanks! $\endgroup$ – Harry Hong Jul 3 '18 at 8:35

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