How to solve $$\lim_{x\to\infty} \left(\frac{4+x}{1+x}\right)^{4+2x}?$$
Well, it looks similar to:
$$e = \lim_{n\to\infty}\left( 1 + n^{-1}\right)^n,$$
but I can't get it solved. Sure, I could use l'Hospital twice (as some calculators suggest). But is that the only way? Isn't there a nice solution that I am too dumb to find?
We have : $$\begin{array}{lcl} \ln \left(\dfrac{4 + x}{1 + x}\right)^{4 + 2x} & = & (4 + 2x) \ln \left(\dfrac{4 + x}{1 + x}\right) \\[3mm] & = & (4 + 2x) \dfrac{\ln \left(\dfrac{4 + x}{1 + x}\right)}{\dfrac{4 + x}{1 + x} - 1} \left(\dfrac{4 + x}{1 + x} - 1\right) \\[3mm] & = & (4 + 2x) \dfrac{\ln \left(\dfrac{4 + x}{1 + x}\right)}{\dfrac{4 + x}{1 + x} - 1} \dfrac{3}{1 + x} \\[3mm] & = & \dfrac{\ln \left(\dfrac{4 + x}{1 + x}\right)}{\dfrac{4 + x}{1 + x} - 1} \dfrac{12 + 6 x}{1 + x} \\[3mm] & \underset{x \to +\infty}{\to} & 1 \times 6 = 6 \end{array}$$ Notice that : $$\lim_{x \to +\infty} \dfrac{\ln \left(\dfrac{4 + x}{1 + x}\right)}{\dfrac{4 + x}{1 + x} - 1} = 1$$ coz : $$\lim_{x \to +\infty} \dfrac{4 + x}{1 + x} = 1$$ and we know that : $$\lim_{t \to 1} \dfrac{\ln t}{t - 1} = 1$$ We deduce that : $$\lim_{x \to +\infty} \left(\dfrac{4 + x}{1 + x}\right)^{4 + 2x} = e^6$$
$$\lim_{x\to\infty} \left(\frac{4+x}{1+x}\right)^{4+2x}=\lim_{x\to\infty} \left(\frac{4+x}{1+x}\right)^{2}\left(\left(1+\frac{3}{1+x}\right)^{\frac{1+x}3}\right)^6=e^6$$
Setting $y=\frac{x+1}{3},$ or $x=3y-1,$ you get the equal limit:
$$\lim_{y\to\infty}\left(1+\frac1y\right)^{2+6y}\tag1$$
Use that $$\lim_{y\to\infty}\left(1+\frac1y\right)^2=1$$ and $$\lim_{y\to\infty} \left(1+\frac1y\right)^y=e$$ to compute $(1).$
$$\left( 1 + \frac{3}{1+x}\right)^{(2+x)^2}=\left(1+\frac{6}{2+2x}\right)^{(2+x)^2}$$
And then I am stuck.
– Korrola Jan 10 '22 at 18:50