Factoring stuff out of limits

**daigo** · June 29th 2012, 03:03 AM

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?

**biffboy** · June 29th 2012, 03:18 AM

Yes you can factor out constants

**emakarov** · June 29th 2012, 03:22 AM

Your derivation is correct. The theorem about the limit of product says that if the two limits on the right exist, then

$\lim\limits_{x \to p} & (f(x)\cdot g(x)) = \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x)$

Since $\lim_{\alpha\to0}\frac{5}{2}$ and $\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}$ exist, your application of the theorem is valid.

**daigo** · June 29th 2012, 03:23 AM

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?

$\lim_{\alpha \rightarrow 0} (\frac{\sin\alpha }{\alpha} \cdot \frac{\sin\alpha }{\alpha}) = (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) \cdot (\lim_{\alpha \rightarrow 0} \frac{\sin\alpha }{\alpha}) = 1 \cdot 1 = 1$

But this wouldn't work with a variable, only constants?

**emakarov** · June 29th 2012, 03:42 AM

The theorem is applicable to the product of any two functions; one of them does not have to be a constant.

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