<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] $$\lim\limits_{n\to \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }=
\lim\limits_{n\to \infty} {\frac{\frac{3^{n+1} + 4}{3^n}\cdot \frac{7^n-47}{7^n}}{\frac{7^{n+1}-47}{7^n}\cdot\frac{3^n +4}{3^n}} }=
\lim\limits_{n\to \infty} {\frac{\left(3+\frac{4}{3^n}\right)\left( 1-\frac{47}{7^n}\right)}{\left(7-\frac{47}{7^n}\right)\left(1+\frac{4}{3^n}\right)} }=
{\frac{\left(\lim\limits_{n\to \infty} 3+\frac{4}{3^n}\right)\left(\lim\limits_{n\to \infty} 1-\frac{47}{7^n}\right)}{\left(\lim\limits_{n\to \infty} 7-\frac{47}{7^n}\right)\left(\lim\limits_{n\to \infty} 1+\frac{4}{3^n}\right)} }=\frac{(3+0)(1+0)}{(7+0)(1+0)}$$
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