Can I solve this limit in this way and why

limitslimits-without-lhopital

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}}+d_m(m^2+2x^2)^{-1} } {\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}}+e_m(m^2+2x^2)^{-1}}
\end{equation}
where
\begin{equation}
\begin{aligned}
m&=1,2,\cdots,M\\
a_m,b,c_m, d_m, e_m &\neq 0 \quad \text{and they are all constants}
\end{aligned}
\end{equation}

My approach:

Since the term $d_m(m^2+2x^2)^{-1}\rightarrow d_m m^{-2}$ and $a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}\rightarrow \infty$, so I ignore the formmer.
Similarly, I ignore the term $e_m(m^2+2x^2)^{-1}$ in the denominator.

Now, the subsequent problem is how to solve the following limit:

\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}
which is solved in my previous question: Is there a mistake in solving this limit?, and @justt and @Claude Leibovici give helpful answers.

My question here is:

  1. Can I ignore the terms $d_m(m^2+2x^2)^{-1}$ and $e_m(m^2+2x^2)^{-1}$ in solving this?

  2. If it is, what is the reason?

I am confused about when can the true value be substituted into the solving of limits.

Sorry about some mistakes in my expressions, as @ancient mathematician points out.

I have rephrased the question.

Best Answer

I think one can argue this way.

Write $A$ for $\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}$, and $C,D,E$ for the other sums.

Claim: Suppose $\lim_{x\to 0}\frac{A}{C}$ exists and equals $\Lambda\ne 0$; then $\lim_{x\to 0}\frac{A+D}{C+E}$ exists and equals $\Lambda$.

To prove this it suffices to prove that $\lim_{x\to 0}\frac{A+D}{C+E}/\frac{A}{C}=1$. As $\frac{A+D}{C+E}/\frac{A}{C}=(1+\frac{D}{A})/(1+\frac{E}{C})$ all we need to see is that $\frac{D}{A}\to 0$ and $\frac{E}{C}\to 0$. As $A,C\to\infty$ it suffices to show that $D,E$ are bounded as $x\to 0$.

We have that $D=\sum_{m=1}^{M}\frac{d_m}{(m^2+2x^2)^{3/2}}$. Let $K$ be the maximum of $|d_1|,\dots, |d_M|$. Then $|D|<KM\frac{1}{(1+2x^2)^{3/2}}\leqslant KM$, so is bounded as required.

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