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

In this question I was able to deduce that the base function $$\left( \frac{x^2+x+1}{3x^2+2x+1} \right) \to 1/3\,\,,\,\,as\,\,x\to\infty $$

But for the power function we have $${\left( \frac{3x^2+x}{x-2} \right)}\to\infty$$ which implies it doesn't exist in $\mathbb R$

so I'm not sure how can we go about solving this case, though the answer seems to be 0.

I've gone through this question which addresses how to deal with cases where we have $ \lim_{x\to c} f(x)^{g(x)} $ but I'm not sure how can it be applied in this case where $ \lim_{x\to \infty} {g(x)} =\infty $.

I also considered taking a log and then computing limits, but for that i guess we need limit to be greater than and $ \lim_{x\to \infty} {g(x)} $ to be a real no.

I'm looking for a bit rigorous solution

grey
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1 Answers1

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You can also use the definition of the limit here to do some more explicit ugly estimations. Assuming these are indeed the limits, you know that for any $M>1$, there exists some $x_M>0$ such that

$$ 0<\frac{x^2+x+1}{3x^2+2x+1} <\frac{1}{2} \quad \text{and} \quad \frac{3x^2+x}{x-2}>M $$

for any $x>x_M$. Then you can say that

$$ 0 \leq \left( \frac{x^2+x+1}{3x^2+2x+1} \right) ^{\left( \frac{3x^2+x}{x-2} \right)} \leq \frac{1}{2^M} $$

for all $x>x_M$.

Keen-ameteur
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  • How did you bound the base function by 1/2? – grey Feb 27 '23 at 15:00
  • Well you said the base function converges to a $1/3$. In particular, from some point onward it should be smaller than a half, or with distance at most $1/6$ from the limit. – Keen-ameteur Feb 27 '23 at 15:11