Why is $\lim_{x\rightarrow \infty}(1+\frac{a}{bx+c})^{dx+f}=e^{\frac{a\cdot d}{b}}$? I am studying limits through a study book, and I am given this simple rule but without any explanation.
$$\lim_{x\rightarrow \infty}(1+\frac{a}{bx+c})^{dx+f}=e^{\frac{a\cdot d}{b}}$$
Why does this hold true for all values of $a, b, c, d, e$ and $f$? This is probably proven by writing the expression in the form of $e^{\ln(...)}$ and then applying l'Hôpital, but I seem to be lost. And secondly, can this also be proven without resorting to l'Hôpital?
Any help would be appreciated, thanks!

Solution
$$\lim_{x\rightarrow \infty}(1+\frac{a}{bx+c})^{dx+f}=\lim_{x\rightarrow \infty}(1+\frac{a}{bx})^{dx}=\lim_{x\rightarrow \infty}(1+\frac{1}{\frac{bx}{a}})^{\frac{bx}{a}\cdot \frac{ad}{b}}$$
Let $u:=\frac{bx}{a}$
$$\lim_{x\rightarrow \infty}(1+\frac{1}{\frac{bx}{a}})^{\frac{bx}{a}\cdot \frac{ad}{b}}=\lim_{u \rightarrow \infty}\left((1+\frac{1}{u})^{u}\right)^{\frac{ad}{b}}=\left(\lim_{u \rightarrow \infty} (1+\frac{1}{u})^u\right)^{\frac{ad}{b}}$$
$\lim_{u \rightarrow \infty} (1+\frac{1}{u})^u$ is the definition of $e$. Here's the why:
$$\lim_{u \rightarrow \infty} (1+\frac{1}{u})^u=e^{\ln(\lim_{u \rightarrow \infty} (1+\frac{1}{u})^u)}=e^{\lim_{u \rightarrow \infty} \left(u \cdot \ln(1+\frac{1}{u})\right)}=e^{\lim_{u \rightarrow \infty} \left(\frac{\ln(1+\frac{1}{u})-\ln 1}{\frac{1}{u}}\right)}=e^{\lim_{\frac{1}{u} \rightarrow 0} \left(\frac{\ln(1+\frac{1}{u})-\ln 1}{\frac{1}{u}}\right)}=e^{(\ln)'(1)}=e$$
Thus:
$$\left(\lim_{u \rightarrow \infty} (1+\frac{1}{u})^u\right)^{\frac{ad}{b}}=e^{\frac{ad}{b}}$$

 A: Assuming $b\ne 0$ we have that
\begin{align}\lim_{x\rightarrow \infty}\left(1+\frac{a}{bx+c}\right)^{dx+f}&\\=&\lim_{x\rightarrow \infty}\left(1+\frac{a}{bx+c}\right)^{{(bx+c)}\cdot\frac{dx+f}{bx+c}}\\=&\left(\lim_{x\rightarrow \infty}\left(1+\frac{a}{bx+c}\right)^{bx+c}\right)^{\lim_{x\to\infty}\frac{dx+f}{bx+c}}.\end{align} Now, use that
$$\lim_{x\rightarrow \infty}\left(1+\frac{a}{bx+c}\right)^{bx+c}=e^a$$ and $$\lim_{x\to\infty}\frac{dx+f}{bx+c}=\dfrac{d}{b}.$$
A: Assume $b\neq 0$.
taking logarithm, we get
$$(dx+f)\ln(1+\frac{a}{bx+c})$$
and using the equivalence
$$\ln(1+X)\sim X\;\;\;(X\to 0)$$
we find
$$\lim_{x\to+\infty}(dx+f)\ln(1+\frac{a}{bx+c})=\lim_{x\to+\infty}\frac{a(dx+f)}{bx+c}=\frac{ad}{b}$$
qed.
A: \begin{align}\left( 1+ \frac{a}{bx+c}\right)^{dx+f} & = e^{(dx+f) \log\left( 1+ \frac{a}{bx+c}\right)}\\
& \approx e^{(dx+f) \left(\frac{a}{bx+c}\right)} \quad \text{since } \log(1+x) \approx x \text{ for small } x \\
& \approx e^{a \times \frac{d+ f/x}{b+c/x}}
\end{align}
As $ x \to \infty$, $$\left( 1+ \frac{a}{bx+c}\right)^{dx+f} \to e^{\frac{ad}{b}}$$
