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

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

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

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, $$F'(x)=F(x)left(lnleft(1+frac{2}{4x-1}right)-frac{8x}{16x^2-1}right)$$ And noting that $frac{mathrm{d}}{mathrm{d}x}left(lnleft(1+frac{2}{4x-1}right)-frac{8x}{16x^2-1}right)=frac{16}{(4x+1)^2(4x-1)^2}$, $$F''(x)=F'(x)left(lnleft(1+frac{2}{4x-1}right)-frac{8x}{16x^2-1}right)+frac{16cdot F(x)}{(4x+1)^2(4x-1)^2}$$ Expanding around $x=0.4$, e.g: $$F(x)approx 1.79774810251-1.05158373234(x-0.4)+6.21731638991(x-0.4)^2+O((x-0.4)^3)$$

Stirling's approximation gives an approximate value for the factorial function $n!$ or the gamma function $Gamma(n)$ when $ngg1$. 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:

$$lim_{xtoinfty}log f(x)=lim_{xtoinfty}logleft(1+frac{2}{4x-1}right)^{x}=lim_{xtoinfty}frac{logleft(1+frac{2}{4x-1}right)}{frac{1}{x}},$$

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

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

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

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

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

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