0
$\begingroup$

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: $\frac{1}{e^2}$

$\endgroup$
  • 1
    $\begingroup$ related: see the most voted answer of math.stackexchange.com/questions/90324/… $\endgroup$ – Quang Hoang Oct 12 '15 at 8:36
  • $\begingroup$ Already upvoted the above comment. One of the tricks taught in schools is to divide and multiply by x inside the parathesis. Forcefully make the limit $e$. and deal with whatever remains. $\endgroup$ – Shailesh Oct 12 '15 at 8:39
1
$\begingroup$

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

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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*}

$\endgroup$
1
$\begingroup$

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 $t \to \infty$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.