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

I tried as follows]1



But at end I got stuck .
$$\lim_{x\to \infty}\left(\frac{ax+1}{bx+2}\right)^x=\lim_{x\to \infty}\left(\frac ab\right)^x\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x \tag 1$$

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 

$$\lim_{x\to \infty} \left(\frac ab\right)^x=\begin{cases}\infty&amp;,a&gt;b&gt;0\\\\0&amp;,b&gt;a&gt;0\\\\1&amp;,a=b&gt;0\end{cases}$$

Therefore, we conclude that if $a&gt;b&gt;0$, then the limit is $\infty$.  If $b&gt;a&gt;0$, then the limit is $0$.  And if $a=b&gt;0$, then the limit is $e^{-1/a}$.
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.