Limits and infinity in a succession

Question

Apologies for this rather basic question. I am preparing the entry exam for university without the help of a teacher and occasionally get stuck on seemingly simple things. I have been all over the internet and cannot find out how to solve this limit:

\begin{align*} \quad \lim_{n \to \infty}\frac{3^n+2^n}{5^n+3}\\ \end{align*}

Does changing it to

\begin{align*} \quad \lim_{n \to \infty}\frac{3^n}{5^n+3} + \lim_{n \to \infty}\frac{2^n}{5^n+3}\\ \end{align*}

would help in this case?

Or can I do

\begin{align*} \quad \lim_{n \to \infty}\frac{1+\frac{2^n}{3^n}}{1+\frac{3}{5^n}}\\ \end{align*}

I would be very, very grateful if anyone could take the time and show me step by step how this is done.

Answer 1

Splitting it into a sum can help. Consider for example

$$ \lim_{n \to \infty} \frac{3^n}{5^n + 3}. $$

Intuitively, as $n \to \infty$, $5^n + 3$ grows much faster than $3^n$ and so you can expect the limit to be zero. To prove it, you can observe that

$$ 0 \leq \frac{3^n}{5^n + 3} \leq \frac{3^n}{5^n} = \left( \frac{3}{5} \right)^n $$

and use the fact that $\lim_{n \to \infty} q^n = 0$ if $|q| < 1$. The second limit can be handled similarly.

Answer 2

Both approaches are valid. Your second one you have made a minor but significant error in. It should be:

$$\lim_{n \to \infty}\frac{1+\frac{2^n}{3^n}}{1+\frac{3}{5^n}}\times\frac{3^n}{5^n}$$

Either will give you the correct answer of zero.