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I want to find the limit as $\sigma_2 \to 0_+$ for $\cfrac{b_1/\sigma_1 + b_2/\sigma_2}{1/\sigma_1 + 1/\sigma_2}$.

By considering $a/\sigma_2$ we notice that as $\sigma_2 \to 0_+$ our expression goes to infinity. Therefore we only need to consider what happens in $\frac{b_2/\sigma_2}{1/\sigma_2}$ a $\sigma_2 \to 0_+$ which is $b_2$ by La'Hopital's rule.


The part that feels too hand wavy is dismissing the $\sigma_1$. I know that's what happens but I don't know how to explain it more rigorously than what I showed above.

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  • $\begingroup$ No need for L'Hopital's rule:$$\frac{\frac{b_1}{\sigma_1}+\frac{b_2}{\sigma_2}}{\frac1{\sigma_1}+\frac1{\sigma_2}}=\frac{\frac{b_1\sigma_2}{\sigma_1}+b_2}{\frac{\sigma_2}{\sigma_1}+1}\to b_2$$ $\endgroup$ – user170231 Sep 30 at 19:54
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There is no problem using L'Hospital:

$$\lim_{\sigma_2\to0}\frac{\dfrac{b_1}{\sigma_1}+\dfrac{b_2}{\sigma_2}}{\dfrac1{\sigma_1}+\dfrac1{\sigma_2}}=\lim_{\sigma_2\to0}\frac{-\dfrac{b_2}{\sigma_2^2}}{-\dfrac1{\sigma_2^2}}=b_2.$$


Without L'Hospital,

$$\lim_{\sigma_2\to0}\frac{\dfrac{b_1}{\sigma_1}+\dfrac{b_2}{\sigma_2}}{\dfrac1{\sigma_1}+\dfrac1{\sigma_2}} =\lim_{\sigma_2\to0}\frac{\dfrac{\sigma_2b_1}{\sigma_1}+b_2}{\dfrac{\sigma_2}{\sigma_1}+1} =\frac{\lim_{\sigma_2\to0}\left(\dfrac{\sigma_2b_1}{\sigma_1}+b_2\right)}{\lim_{\sigma_2\to0}\left(\dfrac{\sigma_2}{\sigma_1}+1\right)}=b_2.$$

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  • $\begingroup$ Could you clarify how you got from the LHS to the RHS in the L'Hospital example? $\endgroup$ – financial_physician Oct 1 at 0:18
  • $\begingroup$ @financial_physician: by applying the formula, what else ? $\endgroup$ – Yves Daoust Oct 1 at 6:33
  • $\begingroup$ What'd you multiply by to go from $\lim_{\sigma_2\to0}\frac{\dfrac{b_1}{\sigma_1}+\dfrac{b_2}{\sigma_2}}{\dfrac1{\sigma_1}+\dfrac1{\sigma_2}}$ to $\lim_{\sigma_2\to0}\frac{-\dfrac{b_2}{\sigma_2^2}}{-\dfrac1{\sigma_2^2}}$ $\endgroup$ – financial_physician Oct 1 at 15:11
  • $\begingroup$ @financial_physician: I didn't multiply, I applied L'Hospital. Do you know what it says ? $\endgroup$ – Yves Daoust Oct 1 at 16:35
  • $\begingroup$ Apparently not haha. What I've seen on it is that you need to have $\pm\infty/\pm\infty$ or $0/0$ and then you can take derivatives, but the first equation doesn't look like that. I see now that you do applied La'Hopital's to get the the equality but it doesn't look like the conditions above. Instead of $0/0$ it looks like undef/undef $\endgroup$ – financial_physician Oct 1 at 17:00
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Since $\frac{1}{\sigma_{1}}$ is a constant and small compared to $\frac{1}{\sigma_{2}}$ as $\sigma_{2}\rightarrow 0^{+}$, we can remove it from the term. Thus we obtain

$$\lim_{\sigma_{2}\rightarrow 0^+}\cfrac{b_1/\sigma_1 + b_2/\sigma_2}{1/\sigma_1 + 1/\sigma_2}$$ $$=\lim_{\sigma_{2}\rightarrow 0^+}\cfrac{b_1/\sigma_1 + b_2/\sigma_2}{ 1/\sigma_2}$$ $$=\lim_{\sigma_{2}\rightarrow 0^+}\cfrac{b_1/\sigma_1}{ 1/\sigma_2} + \frac{b_2/\sigma_2}{ 1/\sigma_2}$$ $$=\lim_{\sigma_{2}\rightarrow 0^+}\cfrac{b_1\sigma_2}{ \sigma_1} + b_2=b_{2}.$$

In fact we also have $$\lim_{\sigma_{2}\rightarrow 0^-}\cfrac{b_1/\sigma_1 + b_2/\sigma_2}{1/\sigma_1 + 1/\sigma_2}=b_2.$$

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  • $\begingroup$ I like the idea of only removing the left hand side of the denominator. Thank you! $\endgroup$ – financial_physician Oct 1 at 0:19
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In this answer I think about $\sigma_2 \to 0_+$ as a sequence of numbers tending towards $0$ from the positive side, and assume that we're comfortable with the fact that the limit of the sum is the sum of the limits.

\begin{align*} \lim_{\sigma_2 \to 0_+} \frac{b_1 / \sigma_1 + b_2 / \sigma_2}{1/\sigma_1 + 1/\sigma_2} &= \lim_{\sigma_2 \to 0_+} \Big( \frac{b_1}{\sigma_1} + \frac{b_2}{\sigma_2} \Big) \frac{\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}\\ &= \lim_{\sigma_2 \to 0_+} \frac{b_1 \sigma_2}{\sigma_1 + \sigma_2} + \lim_{\sigma_2 \to 0_+} \frac{b_2 \sigma_1}{\sigma_1 + \sigma_2}\\ &= \lim_{\sigma_2 \to 0_+} \frac{b_2 \sigma_1}{\sigma_1 + \sigma_2} \text{ (first term certainly goes to zero)}\\ &= b_2. \end{align*}

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