<s>[INST] Calculate: $\lim\limits_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$ How do I calculate the following limit without using l'Hôpital's rule?
$$\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$
 [/INST] $$\dfrac{x^2+2x+3}{x^2+x+1} = 1 + \dfrac{x+2}{x^2+x+1}$$
Hence,
$$\left(\dfrac{x^2+2x+3}{x^2+x+1}\right)^x = \left(1 + \dfrac{x+2}{x^2+x+1} \right)^{\left(\dfrac{x^2+x+1}{x+2} \right) \left(\dfrac{x(x+2)}{x^2+x+1} \right)}$$
Now as $x \to \infty$, we have $\left(1 + \dfrac{x+2}{x^2+x+1} \right)^{\left(\dfrac{x^2+x+1}{x+2} \right)} \to e$ and $\left(\dfrac{x(x+2)}{x^2+x+1} \right) \to 1$.
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