<s>[INST] Limit of $\left(\frac{x^2+x+1}{2x+1}\right)^{1/(x^2-1)}$ when $x\to1$ $$\displaystyle \lim_{x\to 1}\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$
I know that the two-sided limit of $\frac{1}{x^2-1}$ does not exist. I don't know what to do with $\frac{x^2+x+1}{2x+1}$ to get something else than $1^\infty$
 [/INST] Hint
Consider $$A=\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$ Taking logarithms $$\log(A)=\frac{1}{x^2-1}\log\left(\frac{x^2+x+1}{2x+1}\right)$$ Now, consider the Taylor series built at $x=1$ $$\frac{x^2+x+1}{2x+1}=1+\frac{x-1}{3}+\frac{1}{9} (x-1)^2+O\left((x-1)^3\right)$$ and use the fact that, for small $y$, $$\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$$ Now, make $$y=\frac{x-1}{3}+\frac{1}{9} (x-1)^2$$ in order to get the expansion of $\log\left(\frac{x^2+x+1}{2x+1}\right)$ and simplify.
I am sure that you can take from here and conclude not only what is the limit but also how it is approached.
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