MHB Is factoring out of limits allowed?

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    Factoring Limits
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Factoring out limits is permissible when the final limits exist, as demonstrated in the example provided. The transformation of the limit involving sin(α) and α is valid, leading to the conclusion that the limit equals 5/2. Similarly, the manipulation of the limit for sin²(α)/α² is also correct, resulting in a limit of 1. Limit theorems support the distribution of limits across operations like multiplication and division, provided there is no division by zero. Overall, the calculations presented are accurate and adhere to the rules of limits.
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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