Limit problem $\ln(x)$ and $1^\infty$ Can anyone help me with this limit problem without L'Hopital rule and Taylor series?
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}$$
 A: You should definitely compute the limit of the logarithm of your function:
$$
\lim_{x\to1}\frac{\ln(2^x+2)-\ln(3^x+1)}{\ln x}
$$
Substitute $x=t+1$, so the limit becomes
$$
\lim_{t\to0}\frac{\ln(2^{t+1}+2)-\ln(3^{t+1}+1)}{\ln(t+1)}=
\lim_{t\to0}\frac{\ln(2^{t+1}+2)-\ln(3^{t+1}+1)}{t}\frac{t}{\ln(t+1)}
$$
The last factor has limit $1$, so you're left with
$$
\lim_{t\to0}\frac{\ln(2^{t+1}+2)-\ln(3^{t+1}+1)}{t}=
\lim_{t\to0}
\left(\frac{\ln(2^{t+1}+2)-\ln4}{t}-\frac{\ln(3^{t+1}+1)-\ln4}{t}\right)
$$
You should recognize in
$$
\lim_{t\to0}
\frac{\ln(2^{t+1}+2)-\ln4}{t}
$$
and
$$
\lim_{t\to0}
\frac{\ln(3^{t+1}+1)-\ln4}{t}
$$
the derivatives at $0$ of two functions.
If you're not allowed to use derivatives, these are easier to compute anyway.
A: $$\frac{2^x+2}{3^x+1}=\left(1+(\frac{2^x+2}{3^x+1}-1)\right)=1+\frac{2^x-3^3+1}{3^x+1}$$ 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 $$\color{red}{\sqrt[4]{\frac{4}{27}}}$$
A: write your Limit in the form $$e^\lim_{x \to 1}{\frac{\log\left(\frac{2^x+2}{3^x+1}\right)}{\log(x)}}$$
