What is the value of this limit?
- Thread starter radudami
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Hello, and welcome to FMH! 
I would begin by writing:
[MATH]L=\frac{1}{2}\lim_{n\to\infty}\left(\frac{n^{2016}}{\left(\dfrac{3}{2}\right)^n}\right)[/MATH]
Now, consider what will happen after 2016 repeated applications of L'Hôpital's Rule...we would get:
[MATH]L=\frac{2016!}{2\ln^{2016}\left(\dfrac{3}{2}\right)}\lim_{n\to\infty}\left(\frac{1}{\left(\dfrac{3}{2}\right)^n}\right)[/MATH]
Can you proceed?
I would begin by writing:
[MATH]L=\frac{1}{2}\lim_{n\to\infty}\left(\frac{n^{2016}}{\left(\dfrac{3}{2}\right)^n}\right)[/MATH]
Now, consider what will happen after 2016 repeated applications of L'Hôpital's Rule...we would get:
[MATH]L=\frac{2016!}{2\ln^{2016}\left(\dfrac{3}{2}\right)}\lim_{n\to\infty}\left(\frac{1}{\left(\dfrac{3}{2}\right)^n}\right)[/MATH]
Can you proceed?
Once you get it to this point, you see that the base of the exponential form in the denominator is greater than 1.
The numerator is just a polynomial function. You can count on the this limit being 0. Choosing a method to verify
this is another issue.