Compute $\lim_{\sigma_2\to0_+}\frac{b_1/\sigma_1 + b_2/\sigma_2}{1/\sigma_1 + 1/\sigma_2}$. 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.
 A: 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,
$$\lim_{\sigma_2\to0}\frac{\dfrac{b_1}{\sigma_1}+\dfrac{b_2}{\sigma_2}}{\dfrac1{\sigma_1}+\dfrac1{\sigma_2}}
=\lim_{\sigma_2\to0}\frac{\dfrac{\sigma_2b_1}{\sigma_1}+b_2}{\dfrac{\sigma_2}{\sigma_1}+1}
=\frac{\lim_{\sigma_2\to0}\left(\dfrac{\sigma_2b_1}{\sigma_1}+b_2\right)}{\lim_{\sigma_2\to0}\left(\dfrac{\sigma_2}{\sigma_1}+1\right)}=b_2.$$
A: 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 $$\lim_{\sigma_{2}\rightarrow 0^-}\cfrac{b_1/\sigma_1 + b_2/\sigma_2}{1/\sigma_1 + 1/\sigma_2}=b_2.$$
A: 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*}
