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 $1^\infty$
 A: Use :
$$\lim\limits_{x \to 1}h(x)^{g(x)}=\lim\limits_{x \to 1}\mathbb{e}^{g(x)\ln{h(x)}}$$
In your case, $\lim\limits_{x \to 1}g(x)\ln{h(x)}$ is pretty easy :
Using taylor series :
$$\ln h(x) = \frac{(x-1)}{3}+\frac{(x-1)^2}{18} + o((x-1)^2)$$
Then factor a bit :
$$\lim\limits_{x \to 1}g(x)\ln{h(x)}=\lim\limits_{x \to 1}\frac{x+5}{18(x+1)}=\frac{1}{6}$$
A: 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.
A: Let $y=x-1$. Then we have $$\require{cancel}\lim_{y\to0}\left(\frac{y^2+3y+3}{y^2-(y+1)^2+4y+4}\right)^{\frac{1}{y(y+2)}}=\lim_{y\to0}\left(\frac{y^2+3y+3}{2y+3}\right)^{1/(2y)}=\\ \lim_{y\to0}\left(\frac{\cancel{y}(y+3+3/y)}{\cancel{y}(2+3/y)}\right)^{1/(2y)},$$ which, letting $z=1/y$, yields $$\lim_{z\to\infty}\left(\frac{3z+1/z+3}{3z+2}\right)^{z/2}=\lim_{z\to\infty}\left(1+\frac{1+1/z}{3z+2}\right)^{z/2}=\lim_{z\to\infty}\left(1+\frac{1}{3z+2}\right)^{z/2}.$$ Now we could do another substitution, but it is easy to see that $2$ next to $3z$ doesn't count when $z\to\infty$. Thus we can finally conclude the limit is $$\lim_{z\to\infty} \left(1+\frac{1}{3z}\right)^{z/2}=e^{1/6}.$$
