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 $5$. 

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 $y = e^5$
