<s>[INST] Limit problem $\ln(x)$ and $1^\infty$ Can anyone help me with this limit problem without L'Hopital rule and Taylor series?
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}$$
 [/INST] $$\frac{2^x+2}{3^x+1}=\left(1+(\frac{2^x+2}{3^x+1}-1)\right)=1+\frac{2^x-3^3+1}{3^x+1}$$ Puting $\frac{2^x-3^3+1}{3^x+1}=h$ one gets $$\left(\frac{2^x+2}{3^x+1}\right)^{\frac{1}{\ln x}}=(1+h)^{\frac{1}{\ln x}}=(1+h)^{{\frac{h}{h\ln x}}}$$
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}=\lim_{h\rightarrow\ 0}\left((1+h)^{\frac 1h}\right)^{{\frac{h}{\ln x}}}$$ 
Now $$\frac{h}{\ln x}=\frac{2^x-3^x+1}{(3^x+1)\ln x}=\frac{1}{3^x+1}\cdot\frac{2^x-3^x+3-2}{\ln x}$$ it follows
$$\left(\frac{1}{3^x+1}\right) 
 \left(\frac{2(2^{x-1}-1)-3(3^{x-1}-1)}{\ln x}\right)$$ On the other hand, it is known that $$\lim_{x\rightarrow 1}\frac{a^{x-1}-1}{\ln x}=\ln a$$ Hence
$$\lim_{x\rightarrow 1}\left(\frac{1}{3^x+1}\right) 
 \left(\frac{2(2^{x-1}-1)-3(3^{x-1}-1)}{\ln x}\right)=\frac 14(2\ln 2-3\ln 3)$$
Hence we have as limit $$e^{-\frac 14 (\ln 27-\ln 4)}=e^{-\ln \sqrt[4]{\ln 27-\ln 4}}$$ i.e. our limit is equal to $$\color{red}{\sqrt[4]{\frac{4}{27}}}$$
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