rcmango said:
Okay, you after applying l'hopitals rule, i get 1. also, i believe i could change this (n+1)^4 to this n^4 + 1^4 i believe?
thanks.
No, it's not correct at all.
Here's a simple counter example. If n = -1, then your LHS will be 0, whereas your RHS is 2, they are
not equal.
To expand the terms that have the form: (a + b)
n (where n is a natural number), one should use
Binomial Theorem.
Or, you can just divide both numerator, and denominator by n
4 (the greatest power), like this:
\lim_{n \rightarrow \infty} \left| (2x - 1) \frac{n ^ 4 + 16}{(n + 1) ^ 4 + 16} \right|
= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{\frac{n ^ 4 + 16}{n ^ 4}}{\frac{(n + 1) ^ 4 + 16}{n ^ 4}} \right| (since 2x - 1 is a constant, independent of n, we can "pull" it out)
= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\frac{(n + 1) ^ 4}{n ^ 4} + \frac{16}{n ^ 4}} \right|
= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\left( \frac{n + 1}{n} \right) ^ 4 + \frac{16}{n ^ 4}} \right|
= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\left( \frac{n + 1}{n} \right) ^ 4 + \frac{16}{n ^ 4}} \right| = ...
Can you go from here? :)
