# What am I doing wrong finding the limit of $\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$?

$$\lim_{x\to\infty}\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$$ 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?

• $$(1.5)^{-\infty}=?$$ Nov 14, 2018 at 14:27
• @labbhattacharjee ($(\frac{3}{2})^{-\infty}=(\frac{2}{3})^{\infty}=0$, correct? Nov 14, 2018 at 14:30
• @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. Nov 14, 2018 at 14:34 • @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$x^3/(1-x) = x^2/(x^{-1}-1) \ldots$– user284001 Nov 14, 2018 at 14:37 • 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. Nov 14, 2018 at 14:43

Note that

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

but

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

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

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 $$(\frac{3}{2})^{-\infty}=\frac{1}{1.5^{\infty}}=0$$.

• We really don't need that argument to justify the limit.
– user
Nov 14, 2018 at 14:41