how to evaluate a limit similar to Euler identity? [closed]

I have a limit as $x$ approaches infinity of $(1+4/(7x))^x$. Solutions I saw included transforming it into $e$ to the power of natural $\log$ etc. but I've seen a way simpler transformation but can't remember it. Any help appreciated.

If you're familiar with L'Hospital:

$$\lim_{x\to \infty}\ln((1+4/7x)^x)=\lim_{x\to \infty}x\ln(1+4/7x)=\lim_{x\to \infty}\frac{\ln(1+4/7x)}{1/x}=\lim_{x\to \infty}\frac{-4/x(7x+4)}{-1/x^2}=\lim_{x\to \infty}4x/(7x+4)=4/7 $$ This implies: $$\lim_{x\to \infty}(1+4/(7x))^x=e^{\lim_{x\to \infty}\ln((1+4/7x)^x)}=e^{4/7}$$

This doesn't use the property that $\lim_{x\to \infty}(1+t/x)^x=e^t$.

In general for every real $a$, $\lim_{x\to \infty}(1+a/x)^x=e^a$. Indeed, it is the same as $(\lim_{y\to\infty}(1+1/y)^y)^a=e^a$ (replace $x/a$ by $y$). So the answer is $e^{4/7}$.

We have $$\lim_{y\to 0+}\dfrac {\ln (1+y)}{y}=1,$$ because if $y>0$ then $$y=\int_1^{1+y}1\cdot dt>\int_1^{1+y}\frac {1}{t}\cdot dt = \ln (1+y)>\int_1^{1+y}\frac {1}{1+y}\cdot dt=\frac {y}{1+y}.$$

For $x>0$ let $y=4/7x.$ Then $y\to 0^+$ as $x\to\infty$, and $$\ln ((1+4/7x)^x)=x\ln (1+y)=\frac {4}{7} \cdot \frac {\ln (1+y)}{y}.$$