<s>[INST] 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: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$$
 [/INST] 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 $\frac{3}{7}\cdot\frac{1\cdot1}{1\cdot1} = \frac{3}{7}$
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