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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:

$$\lim_{y \to \infty}\left[\frac{y}{\sqrt{1+y^2}} \times \frac{(\frac{1}{y})}{(\frac{1}{y})}\right] =\lim_{y \to \infty} \left[\frac{1}{ \sqrt{\frac{1}{y^2} + 1} }\right]=1$$

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    $\begingroup$ $a\sqrt b=\sqrt {a^{2}b}$ for $ a,b>0$. $\endgroup$ May 15 at 9:49
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    $\begingroup$ I would suggest writing the square roots as fractional powers when you have these sorts of problems, eg $x^\frac{1}{2}$ instead of $\sqrt{x}$. Sometimes this makes things more obvious for me, as you can then apply the power rules that you know already more "mechanically" instead of thinking about the properties of square roots $\endgroup$
    – llama
    May 15 at 20:36

2 Answers 2

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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.

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    $\begingroup$ Yes, since we are taking the limit as $y\to\infty$, we can assume $y$ is greater than any value that we find convenient. $\endgroup$
    – Teepeemm
    May 15 at 19:43
  • $\begingroup$ That is indeed something I haven't noticed but you are completely correct. I took this from my professors notes, and oddly enough, neither had he mentioned what you did here. $\endgroup$
    – Leonardo
    2 days ago
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We have that for $y>0$

$$\sqrt{1+y^2} \times \frac1y =\sqrt{1+y^2} \times \sqrt{ \frac1{y^2} }=\sqrt{1\times \frac1{y^2}+y^2\times \frac1{y^2}}=\sqrt{\frac1{y^2}+1}$$

Note that more in general for $y\neq 0$

$$\sqrt{1+y^2} \times \frac1y =\sqrt{1+y^2} \times \sqrt{ \frac1{y^2} }\times \frac{|y|}{y}=\frac{|y|}{y}\sqrt{\frac1{y^2}+1}$$

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