This might be a basic question but it has been the major issue to solving problems I seem to have when it comes to calculus.
Consider this limit
$$\lim_{\alpha \to \infty}\left[\tanh \alpha\right]= \lim_{y\to \infty}\left[\frac{y}{\sqrt{1+y^2}}\right]=\lim_{y \to \infty}\left[\frac{y}{\sqrt{1+y^2}} \times \frac{(\frac{1}{y})}{(\frac{1}{y})}\right] =\lim_{y \to \infty} \left[\frac{1}{ \sqrt{\frac{1}{y^2} + 1} }\right]=1$$
What is in particular confusing with me is that I always get confused how one manages to "put" $ \frac{1}{y} $ inside the square root to simplify $y^2$.
I think that I know the algebra rules, but this is something that I'm quite sure I haven't encountered so far.
I haven't found any axiom or theorem proving that this is "allowed", although it obviously works.
To make it more legible I'm in particular asking how did we get this to simplify:
$$\lim_{y \to \infty}\left[\frac{y}{\sqrt{1+y^2}} \times \frac{(\frac{1}{y})}{(\frac{1}{y})}\right] =\lim_{y \to \infty} \left[\frac{1}{ \sqrt{\frac{1}{y^2} + 1} }\right]=1$$
Best Answer
The algebra is not so difficult:
$\frac{1}{y}\times\sqrt{1+y^2}=\sqrt{\frac{1}{y^2}}\times\sqrt{1+y^2}$
$=\sqrt{(\frac{1}{y^2})(1+y^2)}$
$=\sqrt{\frac{1}{y^2}+\frac{y^2}{y^2}}=\sqrt{\frac{1}{y^2}+1}$
What needs to be carefully approached is the fact that $\sqrt{y^2}=y$ is only true for $y\geq 0$
I could be wrong here, but I have always assumed that since we are taking the limit as $y\rightarrow \infty$ then $y>0$ is implied.