How to solve $\lim_{x\to\infty} \left(\frac{4+x}{1+x}\right)^{4+2x}$ (without l'Hospital maybe?)

Question

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?

Answer 1

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 $(1).$

Answer 2

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 : $$\lim_{x \to +\infty} \left(\dfrac{4 + x}{1 + x}\right)^{4 + 2x} = e^6$$

Answer 3

$$\lim_{x\to\infty} \left(\frac{4+x}{1+x}\right)^{4+2x}=\lim_{x\to\infty} \left(\frac{4+x}{1+x}\right)^{2}\left(\left(1+\frac{3}{1+x}\right)^{\frac{1+x}3}\right)^6=e^6$$