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$$\lim_{x\to\infty}\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$$ Divide by $x^2$, get $$\lim_{x\to\infty}(1,5)^\infty=\infty$$

The answer in the book is $0$. I've also tried substituting $x$ for $\frac{1}{h}$ where $h$ tends to zero and using some form of $(1+\frac{1}{n})^n$, and using the exponent rule, but everything lands me at infinity.

Do I misunderstand something fundamentally?

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    $\begingroup$ $$(1.5)^{-\infty}=?$$ $\endgroup$ – lab bhattacharjee Nov 14 '18 at 14:27
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    $\begingroup$ @labbhattacharjee ($(\frac{3}{2})^{-\infty}=(\frac{2}{3})^{\infty}=0$, correct? $\endgroup$ – fragileradius Nov 14 '18 at 14:30
  • $\begingroup$ @fragileradius Modulo all of that being essentially gibberish, yes. As far as your question goes, I have no idea what you mean by $(1,5)^\infty$, but it's certainly wrong. At any rate, note that $\frac{x^3}{1-x} < 0 for $x > 1$, and go from there. $\endgroup$ – user3482749 Nov 14 '18 at 14:34
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    $\begingroup$ @user3482749 here $1, 5$ means $1.5 = 3/2$. Commas are often used interchangably with decimal points. the exponent as $\infty$ is wrong however, and should be $- \infty$ as $x^3/(1-x) = x^2/(x^{-1}-1) \ldots$ $\endgroup$ – Kevin Nov 14 '18 at 14:37
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    $\begingroup$ It's not the comma that I'm objecting to. It's the rampant abuse of notation that is using $\infty$ as if it were a real number. $\endgroup$ – user3482749 Nov 14 '18 at 14:43
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Note that

$$\frac{3x^2-x+1}{2x^2+x+1} \to \frac32$$

but

$$\frac{x^3}{1-x}=-\frac{x^3}{x-1}\to -\infty$$

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What you are missing is the fact that $$\lim_{x\to\infty}\frac{x^3}{1-x}\neq \infty$$

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Because the natural logarithm is injective, it turns out that: $\lim_{x \to \infty} \frac{3x²-x+1}{2x²+x+1}^{\frac{x^3}{1-x}}=(\lim_{x \to \infty} \frac{3x²-x+1}{2x²+x+1})^{(\lim_{x \to \infty} \frac{x^3}{1-x})}$. Thus the limit is $(\frac{3}{2})^{-\infty}=\frac{1}{1.5^{\infty}}=0$.

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  • $\begingroup$ We really don't need that argument to justify the limit. $\endgroup$ – user Nov 14 '18 at 14:41

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