Is factoring out of limits allowed?

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    Factoring Limits
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SUMMARY

The discussion confirms that factoring out limits is permissible under certain conditions. Specifically, the user demonstrates the limit calculation of \(\lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\frac{2\alpha}{5}}\) and correctly factors it to obtain \(\frac{5}{2}\). Additionally, the user explores \(\lim_{\alpha \rightarrow 0} \frac{\sin ^{2}\alpha }{\alpha^{2}}\) and verifies that it can be expressed as \((\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha})^{2}\), yielding a result of 1. The discussion emphasizes that these manipulations are valid as long as the final limits exist.

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hatelove
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If I do this:

\lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\frac{2\alpha}{5}} = \lim_{\alpha\rightarrow 0} \frac{5sin\alpha}{2\alpha} = \lim_{\alpha\rightarrow 0} \frac{5}{2}\cdot \frac{sin\alpha}{\alpha}

Am I allowed to do this?

\frac{5}{2} \cdot \lim_{\alpha\rightarrow 0} \frac{sin\alpha}{\alpha} = \frac{5}{2} \cdot 1 = \frac{5}{2}

Or did I do something in the very first steps incorrectly?
 
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Also I didn't want to create another thread since this is kind of relevant:

If I have:

\lim_{\alpha \rightarrow 0} \frac{\sin ^{2}\alpha }{\alpha^{2}} = \lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha})^{2}

Can I do this?

\lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha} \cdot \frac{\sin\alpha }{\alpha}) = (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) \cdot (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) = 1 \cdot 1 = 1
 
You can do all of those things... provided that your final limits exist. They do in this case, so I would say you're on safe ground. There are limit theorems that allow you to distribute the taking of a limit with addition, subtraction, multiplication, division (no divide by zero, of course).
 

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