Oblicz granicę ciągu...- Zadanie 17: MATeMAtyka 3. Zakres podstawowy i rozszerzony

Rozwiązanie

Zapiszmy

$n \to \infty lim \frac{a ^{n}}{3 ^{n} + n - 4 ^{n} - n} = n \to \infty lim \frac{a ^{n} ( 3 ^{n} + n + 4 ^{n} - n )}{( 3 ^{n} + n - 4 ^{n} - n ) ( 3 ^{n} + n + 4 ^{n} - n )} = n \to \infty lim \frac{a ^{n} ( 3 ^{n} + n + 4 ^{n} - n )}{3 ^{n} + n - 4 ^{n} + n} = n \to \infty lim \frac{a ^{n} ( 3 ^{n} + n + 4 ^{n} - n )}{3 ^{n} - 4 ^{n} + 2 n} = n \to \infty lim \frac{a ^{n} \cdot 4 ^{\frac{n}{2}} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{3 ^{n} - 4 ^{n} + 2 n} = n \to \infty lim \frac{a ^{n} \cdot 2 ^{n} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{4 ^{n} ( ( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}} )}$

Dla a = 2⁰ = 1

$n \to \infty lim a_{n} = n \to \infty lim \frac{1 ^{n} \cdot 2 ^{n} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{4 ^{n} ( ( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}} )} = n \to \infty lim \frac{( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}}}{2 ^{n} ( ( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}} )} = \frac{1}{\infty \cdot ( - 1 )} = 0$

Dla a = 2¹ = 2

$n \to \infty lim a_{n} = n \to \infty lim \frac{2 ^{n} \cdot 2 ^{n} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{4 ^{n} ( ( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}} )} = n \to \infty lim \frac{4 ^{n} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{4 ^{n} ( ( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}} )} = n \to \infty lim \frac{( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}}}{( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}}} = \frac{1}{- 1} = - 1$

Dla a = 2² = 4

$n \to \infty lim a_{n} = n \to \infty lim \frac{4 ^{n} \cdot 2 ^{n} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{4 ^{n} ( ( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}} )} = n \to \infty lim \frac{2 ^{n} ( ( \frac{3}{4} ) ^{n} + \frac{n}{4 ^{n}} + 1 - \frac{n}{4 ^{n}} )}{( \frac{3}{4} ) ^{n} - 1 + \frac{2 n}{4 ^{n}}} = \frac{\infty \cdot 1}{- 1} = - \infty$