How can I solve this problem: $\lim_{x\to\infty} (-x +\sqrt{x^2+1})^{1/\ln x}$
Edit: I knew how I had to apply exponentials and logarithms but was unsure of how to work the limits of the exponent. Is this in the picture right?enter image description here Note: I applied L'hopital and despised 1 because of the infinities.
On
Consider $$y=\left(\sqrt{x^2+1}-x\right)^{\frac{1}{\log (x)}}$$ and take logarithms first $$\log(y)=\frac{1}{\log (x)}\log\left(\sqrt{x^2+1}-x\right)$$ Now, consider that $$\sqrt{x^2+1}-x=x\left(\sqrt{1+\frac{1}{x^2}}-1\right)$$ $$\log\left(\sqrt{x^2+1}-x\right)=\log(x)+\log\left(\sqrt{1+\frac{1}{x^2}}-1\right)$$ Now, for small $x$, Taylor series (or generalized binomial expansion) gives $$\sqrt{1+\frac{1}{x^2}}=1+\frac{1}{2 x^2}+O\left(\frac{1}{x^4}\right)$$ $$\sqrt{1+\frac{1}{x^2}}-1=\frac{1}{2 x^2}+O\left(\frac{1}{x^2}\right)$$ $$\log\left(\sqrt{1+\frac{1}{x^2}}-1\right)=-\log(2)-2\log(x)+O\left(\frac{1}{x^2}\right)$$ $$\log\left(\sqrt{x^2+1}-x\right)=-\log(2)-\log(x)+O\left(\frac{1}{x ^2}\right)$$ $$\log(y)=-\frac {\log(x)+\log(2)}{\log(x)}+O\left(\frac{1}{x^2}\right)=-1-\frac {\log(2)}{\log(x)}+O\left(\frac{1}{x^2}\right)$$ which, for $\log(y)$, shows the limit and how it is approached. Now, remember that $y=e^{\log(y)}$.
Note that $$-x+\sqrt{x^2+1}=\frac{1}{x+\sqrt{x^2+1}}$$Then, we can write
$$\begin{align} \lim_{x\to \infty}\left(-x+\sqrt{x^2+1}\right)^{1/\log(x)}&=\lim_{x\to \infty}e^{\frac{-\log\left(x+\sqrt{x^2+1}\right)}{\log(x)}}\\\\ &=\lim_{x\to 0}e^{-\frac{\log(x)+\log\left(1+\sqrt{1+1/x^2}\right)}{\log(x)}}\\\\ &=e^{-1} \lim_{x\to 0}\,e^{-\frac{\log\left(1+\sqrt{1+1/x^2}\right)}{\log(x)}}\\\\ &=e^{-1} \end{align}$$