# Computing limits with complicated polynomials: can we safely drop lower order terms?

I'm computing a limit to find the radius of convergence of a power series function. I've come to the correct answer it appears, but I'm afraid I might have overstepped the rules of limits, my logic feels a bit fuzzy.

Can someone with more experience tell me if this progression was justifiable? Below we join a ratio test for convergence in progress:

$$\lim \limits_{m \to \infty} \left\lvert \frac{xm(3^{m+1}+5m+9)}{(m+1)(3^m+5m+4)} \right\rvert$$ Step 1: Multiply it all out $$\lim \limits_{m \to \infty} \left\lvert \frac{xm3^{m+1}+x5m^2+xm9}{m3^m+3^m+5m^2+9m+4} \right\rvert$$ Step 2: Since $3^m$ will grow faster than $m^2$, that's the high order term, I'm dropping all lower order terms. This is the part I'm least confident in.

$$\lim \limits_{m \to \infty} \left\lvert \frac{xm3^{m+1}}{m3^m+3^m} \right\rvert$$ $$\lim \limits_{m \to \infty} \left\lvert \frac{xm3^{m+1}}{3^m(m+1)} \right\rvert$$ Order the terms for visual benefit $$\lim \limits_{m \to \infty} \left\lvert \frac{x}{} \frac{3^{m+1}}{3^m} \frac{m}{m+1} \right\rvert$$ Now I cancel the $3m$'s leaving a $3$ in the numerator, and the $\frac{m}{m+1}$ because that goes to $1$ as $m$ goes to $\infty$, leaving: $$\lim \limits_{m \to \infty} \left\lvert 3x \right\rvert$$ Giving a radius of convergence for the original power series (not shown) of $\frac{1}{3}$

Did I follow a safe line of reasoning here, or should I have applied a different approach?

Everything seems correct. You're worried about step 2, since you're "dropping" those terms, but what you're actually doing is $$\lim \limits_{m \to \infty} \left\lvert xm3^{m+1}+x5m^2+xm9 \right\rvert = \lim \limits_{m \to \infty} \left\lvert 3^{m+1}(\frac{xm3^{m+1}}{3^{m+1}}+\frac{x5m^2}{3^{m+1}}+\frac{xm9}{3^{m+1}})\right\rvert$$