Solve the limit without using L'Hospital's rule: $\lim _{x\to \infty }\left(\frac{x+2}{x+3}\right)^{2x+3}$

Question

Today here stumbled upon such a limit. He began to solve it, and then I can not understand what to do. $$\lim _{x\to \infty }\left(\frac{x+2}{x+3}\right)^{2x+3}=\lim _{x\to \infty }\left(\:e^{ln\left(\frac{x+2}{x+3}\right)^{2x+3}}\right)=\lim _{x\to \infty }\left(e^{\left(2x+3\right)\cdot \:\:ln\left(\frac{x+2}{x+3}\right)}\right)$$ What's next?

The answer should be: $\frac{1}{e^2}$

Answer 1

Notice that: $$\left(\frac{x+2}{x+3}\right)^{2x+3} = \left(\frac{x+3}{x+3}- \frac{1}{x+3} \right)^{2x+3} = \left(1- \frac{1}{x+3} \right)^{2x+3} = \\ = \left(1- \frac{2}{2x+6} \right)^{2x+6-3}= \left(1- \frac{2}{2x+6} \right)^{2x+6}\left(1- \frac{2}{2x+6} \right)^{-3}$$

Now, perform a change of variable, say $t = 2x+6$. Then:

$$\lim _{x\to \infty }\left(\frac{x+2}{x+3}\right)^{2x+3} = \lim _{t\to \infty }\left(1- \frac{2}{t} \right)^{t}\left(1- \frac{2}{t} \right)^{-3} = \\=e^{-2}(1-0)^{-3} = \frac{1}{e^2}$$

since

$$\lim _{t\to \infty }\left(1+ \frac{k}{t} \right)^{t} = e^k$$

Answer 2

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

Answer 3

The simplest is to start from the standard limit: $$\color{red}{\lim_{u\to+\infty}\Bigl(1+\frac au\Bigr)^{\!u}=\mathrm e^a}.$$ \begin{align*} \left(\frac{x+2}{x+3}\right)^{2x+3}&=\left(1-\frac1{x+3}\right)^{2(x+3)-3}\\ &=\left(\Bigl(1-\frac1{x+3}\Bigr)^{x+3}\right)^2\biggl(1-\frac1{x+3}\biggr)^{-3}\\ &\to(\mathrm e^{-1})^2 1^{-3}=\mathrm e^{-2}. \end{align*}

Answer 4

For $x>0$ let $1/(x+3)=y$. We have $$\lim_{x \to \infty} \left( \frac { x+2}{x+3} \right) ^{2 x+3} =\lim_{y \to 0} (1-y)^{(2 /y)-3}=\lim_{y\to 0} (1-y)^{2 /y}$$ because $ \lim_{y \to 0} (1-y)^{-3}=1. $ Observe that for $0<y<1$ we have $$-y-\frac {y^2/2}{1-y}=-y-\sum_{n=2}^{\infty}y^n/2<-\sum_{n=1}^{\infty}y^n/n=\log (1-y)<-y-y^2/2.$$ Therefore for $0<y<1 $ we have$$(2/y)\left(-y-\frac {y^2/2}{1-y} \right)<\log (1-y)^{2/y}<(2/y)\left(-y-\frac {y^2}{2}\right)$$ and the rest is obvious.

Answer 5

Put $t = x + 3$, then we have $$\left(\frac{x+2}{x+3}\right)^{2x+3} = \left (1 - \frac {1} {t} \right )^{2t - 3} = \left (1 - \frac {1} {t} \right )^{-3} \cdot \left (1 - \frac {1} {t} \right )^{2t} \to 1 \cdot e^{-2} = e^{-2},$$ as $t \to \infty$.