Evaluate: $\lim_{x\to\infty}\left(\frac{ax+b}{cx+d}\right)^x$ I'm trying to evaluate the following limit:
$$
\lim_{x\to\infty}\left(\frac{ax+b}{cx+d}\right)^x
$$
where $a>0$ and $c>0$.
What I tried:
$$
\left(\frac{ax+b}{cx+d}\right)^x = \left(\frac{a+(b/x)}{c+(d/x)}\right)^x = \frac{(a+(b/x))^x}{(c+(d/x))^x} 
$$
Because I wanted to use the following limit:
$$
\lim_{x\to\infty}\left(1+\frac{1}{t}\right
)^t=e
$$
But now I stuck. How to solve it?
 A: You had the right idea. Let me redirect you to the correct way to think.
So, yes you need to use $e$ at some point, but before that you need to get rid of that annoying $x$ in the exponent. An easy way to do that is to apply $\ln$ to your limit, let me show you:
$$\ln(\lim_{x \to \infty} (\frac{ax+b}{cx+d})^x) = \lim_{x \to \infty} \ln((\frac{ax+b}{cx+d})^x) $$
Now, we use the fact:
$$\ln(a^x) = x\ln(a)$$
So, we get:
$$\lim_{x \to \infty} \ln((\frac{ax+b}{cx+d})^x) = \lim_{x \to \infty} x\ln(\frac{ax+b}{cx+d})$$
All we need to solve is $\lim_{x \to \infty} \frac{ax+b}{cx+d} = \lim_{x \to \infty} \frac{x(a+\frac bx)}{x(c+\frac dx)} = \lim_{x \to \infty} \frac{a+\frac bx}{c+\frac dx}  = \frac ac$.
Case 1: $\frac ac \gt 1 \Rightarrow \ln(\frac ac) \gt 0$
Now we can find the limit: $$\lim_{x \to \infty} x\ln(\frac{ax+b}{cx+d}) = \infty \cdot \ln(\frac ac) = + \infty$$
Finally, we shouldn't forget that we applied $\ln$ at the start so we must undo it! To do that, we must apply $e$ to the result since $e^{\ln x} = x$.
So, $$\lim_{x \to \infty} (\frac{ax+b}{cx+d})^x = e^{\infty} = \infty $$

Case 2: $0 \lt \frac ac \lt 1 \Rightarrow \ln(\frac ac) \lt 0$
Now we can find the limit: $$\lim_{x \to \infty} x\ln(\frac{ax+b}{cx+d}) = \infty \cdot \ln(\frac ac) = - \infty$$
Finally, we shouldn't forget that we applied $\ln$ at the start so we must undo it! To do that, we must apply $e$ to the result since $e^{\ln x} = x$.
So, $$\lim_{x \to \infty} (\frac{ax+b}{cx+d})^x = e^{- \infty} = 0 $$

Case 3: $\frac ac = 1 \Rightarrow \ln(\frac ac) = 0$
If we were to find the limit as before we will run into the problem of: $$\lim_{x \to \infty} x\ln(\frac{ax+b}{cx+d}) = \infty \cdot \ln(\frac ac) = \infty \cdot 0$$
an indeterminate form!
To fix that, we can rewrite the last limit to get $\frac 00$, let me show you:
$$\lim_{x \to \infty} x\ln(\frac{ax+b}{cx+d}) = \lim_{x \to \infty} \frac{\ln(\frac{ax+b}{cx+d})}{\frac 1x} = \frac 00$$
$\frac 00$ is useful because we can use L'Hôpital's rule, so let's plug in the derivatives:
$$\lim_{x \to \infty} \frac{\ln(\frac{ax+b}{cx+d})}{\frac 1x} = \lim_{x \to \infty} \frac{\frac{ad-bc}{\left(ax+b\right)\left(cx+d\right)}}{- \frac {1}{x^2}}$$
Now we finally need to do some modifications to finish the limit as follows:
\begin{align}
\lim_{x \to \infty} \frac{\frac{ad-bc}{\left(ax+b\right)\left(cx+d\right)}}{- \frac {1}{x^2}} &= 
\lim_{x \to \infty} \frac{ad-bc}{(ax+b)(cx+d)} \cdot (-x^2)\\ 
&=
\lim_{x \to \infty} \frac{ad-bc}{x(a+\frac bx)x(c+ \frac dx)} \cdot (-x^2) \\
&=
\lim_{x \to \infty} \frac{ad-bc}{(a+\frac bx)(c+ \frac dx)} \cdot (-1) \\
&=
\lim_{x \to \infty} \frac{bc-ad}{(a+\frac bx)(c+ \frac dx)} \\
&=
\frac{bc-ad}{ac} \\
&=
\frac ba - \frac dc
\end{align}
Finally, we shouldn't forget that we applied $\ln$ at the start so we must undo it! To do that, we must apply $e$ to the result since $e^{\ln x} = x$.
So, $$\lim_{x \to \infty} (\frac{ax+b}{cx+d})^x = e^{\frac ba - \frac dc}$$
A: What you need to do is break into cases:-
If $a>c$ . Then $\frac{ax+b}{cx+d}\to\frac{a}{c}>1$ as $x\to\infty$. Hence there exists some $M>0$ such that $x\geq M\implies \frac{ax+b}{cx+d}>r>1$ .
Hence $$\lim_{x\to\infty}(\frac{ax+b}{cx+d})^{x}>\lim_{x\to\infty}r^{x}\,,r>1$$. But the right hand side of this inequality tends to inftty and hence $\displaystyle(\frac{ax+b}{cx+d})^{x}\to\infty$ .
Similarly if $a<c$ you have the correspondingly $x\geq M\implies \frac{ax+b}{cx+d}<r<1$ .
This again gives you that
$$\lim_{x\to\infty}(\frac{ax+b}{cx+d})^{x}<\lim_{x\to\infty}r^{x}\,,r<1$$
And hence $\displaystyle(\frac{ax+b}{cx+d})^{x}\to 0$ .
If $a=c$.
Then we have the $1^{\infty}$ form. When $f(x)^{g(x)}$ have $1^{\infty}$ form we have:-
Then $\displaystyle\lim_{x\to\infty}f(x)^{g(x)}=\exp\bigg(\lim_{x\to\infty }(f(x)-1)g(x)\bigg)$.
This gives you $$\lim_{x\to\infty}(\frac{ax+b}{cx+d})^{x}=\exp\bigg(\lim_{x\to\infty}\frac{(b-d)x}{cx+d}\bigg)=\exp(\frac{b-d}{c})$$
