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I was wondering how can I calculate the limit:

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

without de l'Hôpital rule.

I tried to reconduct the limit at the well known one:

$$\lim_{x\to+\infty}\left(1+\frac1x\right)^x = e$$

Now I'm concentrating only on the part of the limit with still the Indeterminate Form, I reached this form elevating $e$ to the neperian logarithm of the function, trying to get rid of the $1^\infty$ I.F.

but, at the end of the day, I could only obtain:

$$\begin{align}\lim_{x\to+\infty}\ln\left(\frac{1}{x^2}(1+x^2-x)\right) &= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} \ln\left(1 + \frac{1}{\frac{1}{x^2-x}}\right) \\&= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} (x^2-x) \ln\left(\left(1 + \frac{1}{\frac{1}{x^2-x}}\right)^{\frac{1}{x^2-x}}\right)\end{align}$$

But then defining

$$t= \frac{1}{x^2-x}$$

the limit

$$\lim_{t\to 0} \ln\left(\left(1 + \frac{1}{t}\right)^{t}\right)$$

goes no more to

$$ \ln(e)$$

because now $$t \to 0$$

Can you please give me some help?

B. Mehta
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ela
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  • If you are ok, you can accept the answer and set as solved. Thanks! https://math.meta.stackexchange.com/questions/3286/how-do-i-accept-an-answer – user Dec 22 '17 at 13:05

5 Answers5

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$$\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\left(1+\frac{1-x}{x^2}\right)^{\frac{x^2}{1-x}\frac{-3x(1-x)}{2x^2-1}}=\left[\left(1+\frac{1}{\frac{x^2}{1-x}}\right)^{\frac{x^2}{1-x}}\right]^{\frac{3x^2-3x}{2x^2-1}}\to e^\frac32$$

user
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$$ \left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\exp\left[\frac{-3x^3}{2x^2-1}\ln\left(\frac{x^2 -x +1}{x^2}\right)\right]=\exp\left[\frac{-3x^3}{2x^2-1}\ln\left(1-\frac{x-1}{x^2}\right)\right] $$ The exponent $$ \left[\frac{-3x^3}{2x^2-1}\ln\left(1-\frac{x-1}{x^2}\right)\right]\sim-\frac{3}{2}x\left(-\frac{x-1}{x^2}\right)\to 3/2\ , $$ therefore the sought limit is $e^{3/2}$. The above asymptotic estimate is obtained using the Maclaurin expansion of $\ln$.

Pierpaolo Vivo
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If we Put $x=1/t$, the limit becomes $$\lim_{t\to 0^+} (t^2-t+1)^\frac {-3}{t(2-t^2)} $$

Taking logarithm and using the fact that

$$\ln (1-t+t^2)\sim -t+t^2$$

we find that the limit of the logarithm is

$$\lim_{t\to 0^+}\frac {-3}{t (2-t^2)}(-t+t^2)=\frac 32$$

the result is then $$e^\frac 32$$

hamam_Abdallah
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Since $e^x$ and $\ln$ are continuous functions, we obtain: $$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\lim_{x\to+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}\cdot\left(\frac{x^2 -x +1}{x^2}-1\right)\frac{-3x^3}{2x^2-1}}=$$ $$= \lim_{x\to+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}\cdot\frac{3(x^2-x)}{2x^2-1}}=\lim_{x\rightarrow+\infty}e^{\frac{3(x^2-x)}{2x^2-1}\cdot\ln\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}}}=$$ $$=e^{\lim\limits_{x\rightarrow+\infty}\frac{3(x^2-x)}{2x^2-1}\cdot\ln\lim\limits_{x\rightarrow+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}}}=e^{\frac{3}{2}}.$$

Michael Rozenberg
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I thought it might be instructive to present an approach that uses only the squeeze theorem and elementary inequalities obtained using pre-calculus analysis. To that end we proced.

In THIS ANSWER , I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1\tag1$$

Using $(1)$, it is straightforward to see that

$$-\frac{x-1}{x^2-x+1}\le \log\left(1-\frac{x-1}{x^2}\right)\le -\frac{x-1}{x^2}\tag 2$$

Multiplying $(2)$ by $\displaystyle -\frac{3x^3}{2x^2-1}$ and inverting the inequality signs accordingly reveals

$$\begin{align} \underbrace{\left(\frac{3x^3}{2x^2-1}\right)\,\left(\frac{x-1}{x^2}\right)}_{\displaystyle=\frac{3-3/x}{2-1/x^2}}\le -\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)\le \underbrace{ \left(\frac{3x^3}{2x^2-1}\right)\,\left(\frac{x-1}{x^2-x+1}\right)}_{\displaystyle \frac{3-3/x}{(2-1/x^2)(1-1/x+1/x^2)}}\tag 3 \end{align}$$

Applying the squeeze theorem to $(3)$, we find that

$$\lim_{x\to \infty }\left(-\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)\right)=\frac32$$

from which we find by virtue of the continuity of the exponential function that

$$\begin{align} \lim_{x\to \infty }\left(\frac{x^2-x+1}{x^2}\right)^{-\frac{3x^3}{2x^2-1}}&=\lim_{x\to \infty }e^{-\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)}\\\\ &=e^{\lim_{x\to \infty}\left(-\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)\right)}\\\\ &=e^{3/2} \end{align}$$

And we are done!

Main Tools Used: The Squeeze Theorem, Inequalities in $(1)$.

Mark Viola
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