Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$ I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps
\begin{equation}
  \lim_{x\to\infty}  \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
  \end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{x^2+1 - x +1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{x^2 - x +2}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{x \left( x - 1 +\frac{2}{x}\right)}{x \left( \sqrt{1+\frac{1}{x}}+1-\frac{1}{x} \right)}
  \end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{x - 1 +\frac{2}{x}}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
  \end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{\infty - 1 + 0}{1+1-0}
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{\infty - 1}{2} = \infty
\end{equation}
Edit: Added correct steps for completeness, thanks for the quick answers!
\begin{equation}
  \lim_{x\to\infty}  \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{x^2+1 - x^2+2x -1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{2x}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
  \lim_{x\to\infty} \frac{x}{x} \frac{2}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
  \frac{2}{1+1} = 1
\end{equation}

From here we have
$$\frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}=\frac{(\sqrt{x^2+1})^2-(x-1)^2}{\sqrt{x^2+1}+(x-1)}=$$$$=\frac{x^2+1-x^2+2x-1}{\sqrt{x^2+1}+(x-1)}=\frac{2x}{\sqrt{x^2+1}+(x-1)}$$
