taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$$
 A: There are a couple of properties of limits you can abuse here. The one that should pop out first is $\lim f(x)g(x) = \lim f(x) \cdot \lim g(x)$.
$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)}} = \lim_{n\rightarrow \infty} {\frac{3^{n+1}+4}{3^n+4}} \cdot \lim_{n\rightarrow \infty} {\frac{7^n-47}{7^{n+1}-47}} = 3 \cdot \frac{1}{7} = \frac{3}{7}$.
A: You can factor out the exponentials to get: 
$$
\frac{3^{n + 1}7^n\left(1 + \frac{4}{3^{n + 1}}\right)\left(1 - \frac{47}{7^n}\right)}{7^{n+1}3^n\left(1 - \frac{47}{7^{n + 1}}\right)\left(1 + \frac{4}{3^n}\right)}
$$
Which leaves us with:
$$
\frac{3}{7}\cdot\frac{\left(1 + \frac{4}{3^{n + 1}}\right)\left(1 - \frac{47}{7^n}\right)}{\left(1 - \frac{47}{7^{n + 1}}\right)\left(1 + \frac{4}{3^n}\right)}
$$
Now when you take the limit as $n$ goes to $\infty$, you simply end up with $\frac{3}{7}\cdot\frac{1\cdot1}{1\cdot1} = \frac{3}{7}$
A: $$\lim\limits_{n\to \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }=
\lim\limits_{n\to \infty} {\frac{\frac{3^{n+1} + 4}{3^n}\cdot \frac{7^n-47}{7^n}}{\frac{7^{n+1}-47}{7^n}\cdot\frac{3^n +4}{3^n}} }=
\lim\limits_{n\to \infty} {\frac{\left(3+\frac{4}{3^n}\right)\left( 1-\frac{47}{7^n}\right)}{\left(7-\frac{47}{7^n}\right)\left(1+\frac{4}{3^n}\right)} }=
{\frac{\left(\lim\limits_{n\to \infty} 3+\frac{4}{3^n}\right)\left(\lim\limits_{n\to \infty} 1-\frac{47}{7^n}\right)}{\left(\lim\limits_{n\to \infty} 7-\frac{47}{7^n}\right)\left(\lim\limits_{n\to \infty} 1+\frac{4}{3^n}\right)} }=\frac{(3+0)(1+0)}{(7+0)(1+0)}$$
