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I am trying to calculate $$\lim _{x\to \infty }\left(\frac{x^2+3x+2}{x^2\:-x\:+\:1\:}\right)^x$$

My initial thought is that it is in exponential form $\left(1+\frac{a}{f(x)}\right)^{f(x)}$.

I tried to factor the polynomials $\frac{(x+1)(x+2)}{x(x-1)+1}$ in order to bring it to that form, but had no success.

I also tried to apply the chain rule as following, but found nothing interesting either

$$e^{x\ln({\left(\frac{x^2+3x+2}{x^2\:-x\:+\:1\:}\right)}}$$

Any ideas that don't involve D'Hopital's Rule?

HelloWorld
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    You should write the fraction as $$1+\frac{4x+1}{x^2-x+1}$$, then use the limit definition of the exponential function – Yuriy S Jul 09 '20 at 10:57

4 Answers4

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The logarithm is $\lim_{x\to\infty}x\ln\left(1+\tfrac{4x+1}{x^2-x+1}\right)$. As $x\to\infty$, $\tfrac{4x+1}{x^2-x+1}\sim\tfrac4x\to0$, and since $\lim_{y\to0}\tfrac{\ln(1+y)}{y}=:\ln^\prime1=1$ we have$$\lim_{x\to\infty}x\ln\left(1+\tfrac{4x+1}{x^2-x+1}\right)=\lim_{x\to\infty}\tfrac{x(4x+1)}{x^2-x+1}=4.$$

J.G.
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It's $$\left(1+\frac4x+O(x^{-2})\right)^x$$ whose logarithm is $$x\left(\frac4x+O(x^{-2)}\right)=4+O(x^{-1}).$$ So the original limit is $e^4$.

Angina Seng
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Your initial thought is right. Keep thinking. $$\lim _{x\to \infty }\left(\frac{x^2+3x+2}{x^2\:-x\:+\:1\:}\right)^x=\lim _{x\to \infty }\left(1+\frac{4x+1}{x^2\:-x\:+\:1\:}\right)^x=\\ \lim _{x\to \infty }\left[\left(1+\frac{1}{f(x)}\right)^{f(x)}\right]^{\frac{x}{f(x)}}=\exp\left({\lim\limits_{x\to \infty }\frac{4x^2+x}{x^2\:-x\:+\:1\:}}\right)=e^4.$$ where $f(x)=\frac{x^2-x+1}{4x+1}$ and $\lim_\limits{x\to\infty} f(x)=\infty$.

farruhota
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This limit is is the form $1^{\infty}$. We will use the formula:

$$\lim_{x\to a} f^g=e^{\lim_{x\to a}g(f-1)}$$

(See link)

The rest is pretty straight forward: $$g(f-1)\equiv x\left(\frac{4x+1}{x^2-x+1}\right)\to 4$$ When $x\to \infty$.

Thus the limit is $e^4$.

AryanSonwatikar
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