# 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.

• No need for L'Hopital's rule:$$\frac{\frac{b_1}{\sigma_1}+\frac{b_2}{\sigma_2}}{\frac1{\sigma_1}+\frac1{\sigma_2}}=\frac{\frac{b_1\sigma_2}{\sigma_1}+b_2}{\frac{\sigma_2}{\sigma_1}+1}\to b_2$$ – user170231 Sep 30 at 19:54

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,

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

• Could you clarify how you got from the LHS to the RHS in the L'Hospital example? – financial_physician Oct 1 at 0:18
• @financial_physician: by applying the formula, what else ? – Yves Daoust Oct 1 at 6:33
• 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 $\lim_{\sigma_2\to0}\frac{-\dfrac{b_2}{\sigma_2^2}}{-\dfrac1{\sigma_2^2}}$ – financial_physician Oct 1 at 15:11
• @financial_physician: I didn't multiply, I applied L'Hospital. Do you know what it says ? – Yves Daoust Oct 1 at 16:35
• 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 – financial_physician Oct 1 at 17:00

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

• I like the idea of only removing the left hand side of the denominator. Thank you! – financial_physician Oct 1 at 0:19

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*}