How can I approximate $ \left(1+\frac{2}{4x-1}\right)^{x} $

Question

I want to approximate

$$ f(x) = \left(1+\frac{2}{4x-1}\right)^{x} $$

I begin with the Stirling Approximation

$$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^{n}$$

Raise both sides to the power of $$ \frac{n+1}{4n(n-1)} = A $$

$$(n!)^{A} \sim \left(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^{n}\right)^{A} $$

When I plot this on Desmos it looks fine , However if i let :

$$ n = \left(1+\frac{2}{4x-1}\right )$$

To obtain the desired function

$$ \left(1+\frac{2}{4x-1}\right)^{x} $$

on the right side it does not look Asymptotic as $ x \to \infty $

What can i do , or what function do i use to approximate $ f(x) $ as $x \to \infty$

Thank you very much for your help and time.

Answer 1

Are you looking for $\lim _{x\ \rightarrow \infty \ }\left(1+\frac{2}{4x-1}\right)^x\ $?

If so, $$\lim _{x\ \rightarrow \infty \ }\left(1+\frac{2}{4x-1}\right)^x\ =\lim _{x\ \rightarrow \ \infty \ }\left(1+\frac{1}{\frac{4x-1}{2}}\right)^{\frac{4x-1}{2}\cdot \frac{2x}{4x-1}}=\lim _{x\ \rightarrow \ \infty \ }\left(\left(1+\frac{1}{\frac{4x-1}{2}}\right)^{\frac{4x-1}{2}}\right)^{\lim _{x\ \rightarrow \ \infty \ }\ \frac{2x}{4x-1}}=e^{\frac{1}{2}}=\sqrt{e}.$$

Answer 2

Stirling's approximation gives an approximate value for the factorial function $n!$ or the gamma function $\Gamma(n)$ when $n\gg1$. In your case, there are no factorial functions or gamma functions in $f(x)$ so there is no need to use Stirling's approximation. Instead, take the logarithm:

$$\lim_{x\to\infty}\log f(x)=\lim_{x\to\infty}\log\left(1+\frac{2}{4x-1}\right)^{x}=\lim_{x\to\infty}\frac{\log\left(1+\frac{2}{4x-1}\right)}{\frac{1}{x}},$$

and notice that this limit will approach $0/0$. Therefore we can apply L'Hopital's rule. The derivatives of the numerator and denominator are

$$\frac{d}{dx}\log\left(1+\frac{2}{4x-1}\right)=\frac{8}{1-16x^2},\quad\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{-1}{x^2},$$ so the limit may be rewritten as

$$\lim_{x\to\infty}\frac{8x^2}{16x^2-1}\to \frac{1}{2}.$$

Can you finish the problem and find $\underset{x\to\infty}\lim f(x)$?

Answer 3

Why not use a Taylor expansion? We can use the identity $$\frac{\mathrm{d}}{\mathrm{d} x}\left(f(x)^{g(x)}\right)=f(x)^{g(x)}\left(g'(x)\ln(f(x))+\frac{f'(x)}{f(x)}g(x)\right)$$ Let $F(x)=\left(1+\frac{2}{4x-1}\right)^x$. Then, $$F(x)\approx F(a)+F'(a)(x-a)+\frac{F''(a)(x-a)^2}{2}+O((x-a)^3)$$ In the neighborhood of $x=a$. We can compute using the above identity, $$F'(x)=F(x)\left(\ln\left(1+\frac{2}{4x-1}\right)-\frac{8x}{16x^2-1}\right)$$ And noting that $\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln\left(1+\frac{2}{4x-1}\right)-\frac{8x}{16x^2-1}\right)=\frac{16}{(4x+1)^2(4x-1)^2}$, $$F''(x)=F'(x)\left(\ln\left(1+\frac{2}{4x-1}\right)-\frac{8x}{16x^2-1}\right)+\frac{16\cdot F(x)}{(4x+1)^2(4x-1)^2}$$ Expanding around $x=0.4$, e.g: $$F(x)\approx 1.79774810251-1.05158373234(x-0.4)+6.21731638991(x-0.4)^2+O((x-0.4)^3)$$

Answer 4

As, $(1+\frac{2}{4x-1})^{x} = e^{\ln(1+\frac{2}{4x-1})^{x}} = e^{x \ln(1+\frac{2}{4x-1})} $.

Now, $$\lim_{x \to \infty} (1+\frac{2}{4x-1})^{x} $$ $$=\lim_{x \to \infty} e^{x \ln(1+\frac{2}{4x-1})}$$ $$=e^{\lim_{x \to \infty} \frac{\ln(1+\frac{2}{4x-1})}{\frac{1}{x}}}$$ $$=e^{\lim_{x \to \infty} \frac{\frac{-2}{(4x-1)^2} \cdot 4 }{(1+\frac{2}{4x-1}) \cdot \frac{-1}{x^2}}} (\text{using L'hospital rule}) $$ $$=e^{\lim_{x \to \infty} \frac{\frac{8}{(4-\frac{1}{x})^2}}{(1+\frac{2}{4x-1})}} $$ $$=e^{\frac{8}{4^2}} $$ $$=\sqrt{e} $$