How to calculate $\lim_{n\rightarrow\infty}\frac{a(a+1)(a+2)\cdots(a+n)}{b(b+1)(b+2)\cdots(b+n)}$ How to calculate $$\lim_{n\rightarrow\infty}\frac{a(a+1)(a+2)\cdots(a+n)}{b(b+1)(b+2)\cdots(b+n)}$$
where $a>0, b>0?$
I could not calculate the limit with ordinary tool(ratio test, squeeze ...).
Does anyone have an idea of efficiency to find this limit?
Thank you!

An approach that doesn't use asymptotics:
We consider the case $a>r$, that is, $a = b + r$ with $r>0$.  Our sequence satisfies the recurrence
$$
f(n) = f(n-1)\frac{b+r+n}{b+n} = f(n-1)\left(1 + \frac{r}{b+n} \right)
$$
We note that our sequence is positive and monotonically increasing.  It follows that
$$
f(n) - f(n-1) = f(n-1)\left(1 + \frac{r}{b+n} \right) - f(n-1) =\\
\frac{r}{b + n}f(n-1) >
\frac{r}{b + n}f(0) = \frac{a}{b}\frac{r}{b+n}
$$
We then note that
$$
\lim_{n \to \infty} f(n) = f(0) + 
\lim_{n \to \infty} \sum_{k=1}^n (f(k) - f(k-1)) >
f(0) + f(0)\lim_{n \to \infty} \sum_{k=1}^n \frac{r}{b + k}
$$
By the divergence of the sum, we conclude that the limit is $\infty$.

In the case $a = b$, note that $f(n) = 1$, so our limit is $1$.

In the case $a < b$, our limit can be rewritten as
$$
\lim_{n\rightarrow\infty}\left(\frac{b(b+1)(b+2)\cdots(b+n)}{a(a+1)(a+2)\cdots(a+n)}\right)^{-1}
$$
so that from our previous analysis, the limit is $0$.
