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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:

  1. Is there a mistake in my derivation?

  2. If I made mistakes in the derivation, then, what is the correct derivation?

Thanks for helpful comments and answers!

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2 Answers 2

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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.

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Since @justt gave the answer, just try.

Take an example : $a_m=m$, $c_m=m^2$, $b=1$, $M=4$.

Without any simplification at all, the limit is $$\frac{144+36 e^{3/2}+16 e^{16/9}+9 e^{15/8}}{144+72 e^{3/2}+48 e^{16/9}+36 e^{15/8}}=0.465476$$

Ignoring the term $2x^4$, the limit becomes $\frac{41}{60}$

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