Limit of indeteminate form $1^{(∞)}$ If we consider the function $f(x)=[(ax+1)/(bx+2)]^{x}$ where $a,\,b >0$ and $a<b,\,$ then we have to find $\lim_{x\to \infty} f(x)$ 
I tried as follows]1

But at end I got stuck .
 A: $$\lim_{x\to \infty}\left(\frac{ax+1}{bx+2}\right)^x=\lim_{x\to \infty}\left(\frac ab\right)^x\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x \tag 1$$
Using the limit definition of the exponential function, we see that
$$\lim_{x\to \infty}\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x=e^{1/a-2/b}$$
However, we have 
$$\lim_{x\to \infty} \left(\frac ab\right)^x=\begin{cases}\infty&,a>b>0\\\\0&,b>a>0\\\\1&,a=b>0\end{cases}$$
Therefore, we conclude that if $a>b>0$, then the limit is $\infty$.  If $b>a>0$, then the limit is $0$.  And if $a=b>0$, then the limit is $e^{-1/a}$.
A: Equations with messy powers are usually dealt with better by taking logs of everything in sight.
So instead of looking for $\lim_{x \to \infty} f(x)$, look for $\lim_{x \to \infty} log(f(x))$  Once you know that, you can easily find the original limit easily.
It is easy to start.
$$\begin{align}\log(f(x)) &= \log((\frac{ax + 1}{bx + 2})^x) \\
&= x \log(\frac{ax + 1}{bx + 2}) \\
&= x (log(ax + 1) - log(bx + 2))\end{align}$$.
Now we'd prefer to see something like $log(ax) - log(bx)$ but we have something messier.  So let's rearrange to get that and a hopefully small error term.
$$\begin{align}\log(f(x)) &= x (log(ax + 1) - log(bx + 2))\\
&= x (log((ax + 1)\frac{ax}{ax} - log((bx+2)\frac{bx}{bx})\\
&= x (log(ax(1 + \frac{1}{ax})) - log(bx(1 + \frac{2}{bx})))\\
&= x (log(a) + log(x) + log(1 + \frac{1}{ax}) - (log(b) + log(x) + log(1 + \frac{2}{bx})))\\
&= x (log(a/b) + log(1 + \frac{1}{ax}) - log(1 + \frac{2}{bx}))\end{align}$$
And now you have 3 cases.  $a<b$, $a=b$ and $a>b$.  The first and third have obvious answers.  For the complicated middle case, you can expand the Taylor series for $\log$ around $1$.
That gives you the limit of $\log(f(x))$.  Undo the log, and you get your answer.
A: In any situation with this indeterminant form, I always use log and L'Hopital. First, consider $\lim\log (f (x)) $ instead. We're now looking at $\lim x(\log (ax + 1) - \log (bx + 2))$, after applying a couple of log properties. Next, rephrase it as $\lim\frac {\log (ax +1) - \log (bx +2)}{1/x} $. This has form 0/0, so L'Hopital's rule applies.
