I have been trying to solve this limit:
$\lim \limits_{n \to \infty} \sqrt{16^n - 4^n} - \sqrt{16^n - 3^n + n} $
The limit should be - 1/2. I have been trying to use Squeeze Theorem but no luck trying to find the lower bound with limit - 1/2.
Any help would be appreciated!
HINT:
By rationalization, $$\dfrac{16^n-4^n-(16^n-3^n+n)}{\sqrt{16^n-4^n}+\sqrt{16^n-3^n+n}}$$
$$=\dfrac{(3/4)^n-1-n/4^n}{\sqrt{1-(4/16)^n}+\sqrt{1-(3/16)^n+n/16^n}}$$
Use How to prove that $\lim_{n \to \infty} n x^{n} = 0 $ when $0<x<1$?
$$\sqrt{16^n - 4^n} - \sqrt{16^n - 3^n + n}=\frac{3^n-4^n-n}{\sqrt{16^n - 4^n} + \sqrt{16^n - 3^n + n}}=$$
$$=\require{cancel}\frac{\cancel{4^n}\left(\left(\frac34\right)^n-1-\frac n{4^n}\right)}{\cancel{4^n}\left(\sqrt{1-\left(\frac14\right)^n}+\sqrt{1-\left(\frac3{16}\right)^n+\frac n{16^n}}\right)}\xrightarrow[n\to\infty]{}\frac{0-1-0}{\sqrt1+\sqrt1}=-\frac12$$
$$\quad \lim _{ n\to \infty } \left( \sqrt { 16^{ n }-4^{ n } } -\sqrt { 16^{ n }-3^{ n }+n } \right) =\lim _{ n\to \infty } \frac { \left( \sqrt { 16^{ n }-4^{ n } } -\sqrt { 16^{ n }-3^{ n }+n } \right) \cdot \left( \sqrt { 16^{ n }-4^{ n } } +\sqrt { 16^{ n }-3^{ n }+n } \right) }{ \left( \sqrt { 16^{ n }-4^{ n } } +\sqrt { 16^{ n }-3^{ n }+n } \right) } =\\ =\lim _{ n\to \infty } \frac { 16^{ n }-4^{ n }-16^{ n }+3^{ n }-n }{ \left( \sqrt { 16^{ n }-4^{ n } } +\sqrt { 16^{ n }-3^{ n }+n } \right) } =\lim _{ n\to \infty } \frac { -4^{ n }+3^{ n }-n }{ \left( \sqrt { 16^{ n }-4^{ n } } +\sqrt { 16^{ n }-3^{ n }+n } \right) } =\lim _{ n\to \infty } \frac { { 4 }^{ n }\left( { \left( \frac { 3 }{ 4 } \right) }^{ n }-1-\frac { n }{ { 4 }^{ n } } \right) }{ { 4 }^{ n }\left( \sqrt { 1-\frac { 1 }{ { 4 }^{ n } } } +\sqrt { 1-{ \left( \frac { 3 }{ 4 } \right) }^{ n }+\frac { n }{ { 4 }^{ n } } } \right) } =\\ =\frac { 0-1-0 }{ 1+1+0 } =-\frac { 1 }{ 2 } $$
