I want to solve this:
\begin{equation}
L=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}}
\end{equation}
where
\begin{equation}
\begin{aligned}
m=1,2,\cdots,M\\
a_m,b,c_m \neq 0 \quad \text{and are all constants}
\end{aligned}
\end{equation}
Here is my approach:
Since $x\rightarrow 0$, I ignore the term $2x^4$ in the denominator of $\exp\left\{\frac{b_m}{ (m^2+2x^2)x^2}\right\}$ and I get:
\begin{equation}
\begin{aligned}
L&=\lim_{x\rightarrow 0} \frac{\exp\left\{\frac{b}{x^2}\right\}\sum\limits_{m=1}^{M}a_{m}{(m^2+2x^2)^{-3/2}} } {\exp\left\{\frac{b}{x^2}\right\}\sum\limits_{m=1}^{M} c_{m}{(m^2+2x^2)^{-3/2}}}\\
&=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}{(m^2+2x^2)^{-3/2}}}\\
&=\frac{\sum\limits_{m=1}^{M}\frac{a_{m}}{m^3} } {\sum\limits_{m=1}^{M} \frac{c_{m}}{m^3}}
\end{aligned}
\end{equation}
My question is:
-
Is there a mistake in my derivation?
-
If I made mistakes in the derivation, then, what is the correct derivation?
Thanks for helpful comments and answers!
Best Answer
If $x^2 \to 0$ then you certainly can't say that $\exp(\frac{b}{x^2}) \to 1$. Quite the contrary, this term goes to infinity very fast, the larger $b$ the faster.
You started with neglecting a smaller order term, that's all and well to get a rough idea, but that's certainly not rigorous, all the more inside an exponential. So let's factor out that $\exp(b/x^2)$ growth rate and see what's left.
$\exp(\frac{bm^2}{ (m^2+2x^2)x^2}) = \exp(\frac{b}{x^2})\exp(\frac{bm^2}{ (m^2+2x^2)x^2} - \frac{b}{x^2}) = \exp(\frac{b}{x^2})\exp(\frac{bm^2-b(m^2+2x^2)}{ (m^2+2x^2)x^2} ) = \exp(\frac{b}{x^2})\exp(\frac{-2b}{m^2+2x^2}). $
Now you can plug that in your sum and cancel out the $\exp(\frac{b}{x^2})$'s. You will end up with something nicely convergent.