Is there a mistake in solving this limit?

I want to solve this: $$$$L=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}}$$$$ where \begin{aligned} m=1,2,\cdots,M\\ a_m,b,c_m \neq 0 \quad \text{and are all constants} \end{aligned}

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{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}

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?

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.

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 $$\frac{41}{60}$$