Evaluating $\lim_{x\to\infty}\left(\frac{2+x}{-3+x}\right)^{x}=e^5$ 
Show That$$\lim_{x\to\infty}\left(\frac{2+x}{-3+x}\right)^{x}=e^5$$

How does one reach this result? I keep getting $+3$ in the power and not only $5$. 
 A: Another possible way could use Taylor expansion $$A=\left(\frac{x+2}{x-3}\right)^x=\left(1+\frac{5}{x-3}\right)^x$$ $$\log(A)=x\log\left(1+\frac{5}{x-3}\right)$$ Now, remembering that $$\log(1+y)=y+O\left(y^2\right)$$ replace $y$ by $\frac{5}{x-3}$; so, $$\log(A)\approx \frac{5x}{x-3}$$ and $$A\approx e^{\frac{5x}{x-3}}$$ and then the result.
Going further in the expansion for large values of $x$, you could show that the asymptotic expansion looks like $$A=e^5\Big(1+\frac{5 }{2 x}\Big)+O\left(\left(\frac{1}{x}\right)^2\right)$$ which reveals also how is reached the limit.
A: The limit is:
$$y = \lim_{x\to\infty} \left(\frac{x+2}{x-3}\right)^x$$
Taking the natural log of both sides:
$$\ln y = \lim_{x\to\infty} x\ln\left(\frac{x+2}{x-3}\right) $$
Notice that:
$$\ln y = \lim_{x\to\infty} x\ln\left(\frac{x+2}{x-3}\right) = \lim_{x\to\infty}\frac{\ln\left(\frac{x+2}{x-3}\right)}{\frac{1}{x}}$$
Direct substitution yields $\frac{0}{0}$. Using L'Hospital's Rule:
$$\lim_{x\to\infty}\frac{\ln\left(\frac{x+2}{x-3}\right)}{\frac{1}{x}} = \frac{\frac{5}{-x^2+x+6}}{-\frac{1}{x^2}} = \frac{5x^2}{x^2 -x +6}$$
Now, remember the rule:
$$\lim_{x\to\infty}\frac{5x^2}{x^2 -x +6} = \frac{\frac{5x^2}{x^2}}{\frac{x^2}{x^2}-\frac{x}{x^2}+\frac{6}{x^2}}$$
Taking the limit we get $5/1 = 5$
Now, $\ln y = 5$
Therefore, we get $y = e^5$
A: $$\begin{align}\lim_{x\to\infty}\left(\frac{2+x}{-3+x}\right)^x&=\lim_{x\to\infty}\left[\left(1+\frac{5}{-3+x}\right)^{\frac{-3+x}{-5}}\right]^{\frac{5x}{-3+X}}\\&=e^{\lim_{x\to\infty}\frac{5x}{-3+x}}\\&=e^{5}\end{align}$$
A: Hint:
$$\lim_{x\to\infty} \Bigg(\frac{x - 3 + 5}{x - 3}\Bigg)^{\frac{5x}{x-3}\frac{x-3}{5}} = \lim_{x\to\infty} \Bigg[\Bigg(1 + \frac{ 5}{x - 3}\Bigg)^{\frac{x-3}{5}}\Bigg]^{\frac{5x}{x-3}}$$
And use $\lim_{x\to \infty} (1 + \frac{1}{x})^x = e$.
A: I would consider a substitution and use the limit definiton of $e^x$. Say $x=y+3$ and $y\to \infty$. We can than say $$\lim_{y\to\infty}\left(\frac{y+5}{y}\right)^{y+3}=\lim_{y\to\infty}\left(1+\frac5y\right)^{y+3}=\lim_{y\to\infty}\underset{=e^5}{\underbrace{\left(1+\frac5y\right)^y}}\cdot\underset{=1}{\underbrace{\left(1+\frac5y\right)^3}}=e^5.$$
A: Taking $\log$ and using the equivalence $\log(1+f(x))\approx f(x)$ when $f(x)\to 0$:
$$
\log L=\lim_{x\to\infty} x\ln\left(\frac{x+2}{x-3}\right)=
\lim_{x\to\infty}\frac{\ln\left(\frac{x+2}{x-3}\right)}{\frac{1}{x}}
=\lim_{x\to\infty}\frac{\left(\frac{x+2}{x-3}\right)-1}{\frac{1}{x}}=\cdots
$$
