Calculate: $\lim\limits_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$

$$\dfrac{x^2+2x+3}{x^2+x+1} = 1 + \dfrac{x+2}{x^2+x+1}$$ 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 $\left(\dfrac{x(x+2)}{x^2+x+1} \right) \to 1$.


A slightly different take. Let $L$ be the limit in question. Then we have

$$\begin{align}\log{L} &= \lim_{x \rightarrow \infty}x \log{\left(\frac{x^2+2 x+3}{x^2+x+1}\right)}\\ &= \lim_{x \rightarrow \infty}x \log{\left( 1+\frac{x+2}{x^2+x+1}\right)}\\ &= \lim_{x \rightarrow \infty} \frac{x(x+2)}{x^2+x+1}\end{align}$$

Therefore $\log{L}=1$ and $L=e$.


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

$$=\lim_{x \rightarrow \infty}\left(1+\frac{x+2}{x^2+x+1} \right)^x$$

$$=\lim_{x \rightarrow \infty}\left(\left(1+\frac{x+2}{x^2+x+1} \right)^\frac{x^2+x+1}{x+2}\right)^{\frac{x(x+2)}{x^2+x+1}}$$

$$=e$$ as $\lim_{x\to\infty}\frac{x(x+2)}{x^2+x+1}=\lim_{x\to\infty}\frac{(1+2/x)}{1+1/x+1/{x^2}}=1$

and $\lim_{x\to\infty}\left(1+\frac{x+2}{x^2+x+1} \right)^\frac{x^2+x+1}{x+2}=\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$

Tags:

Limits

Calculus

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