Limity s parametry v exponentu
Úloha číslo: 2843
Určete limity v závislosti na parametrech \(k,l\in\mathbb N\):
Varianta 1
\(\displaystyle \lim_{n\to\infty} \frac{n^k-(n-1)^l}{n^k+n^l}\)
Řešení
Pro \(k>l\) vytkneme \(n^k\):
\(\displaystyle \lim_{n\to\infty} \frac{n^k-(n-1)^l}{n^k+n^l}= \lim_{n\to\infty} \frac{1-\left(\frac{n-1}n\right)^l n^{l-k}}{1+n^{l-k}}= \frac{1-(1-0)^l \cdot 0}{1+0}=1. \)
Pro \(k=l\) vytkneme \(n^k\):
\(\displaystyle \lim_{n\to\infty} \frac{n^k-(n-1)^k}{n^k+n^k}= \lim_{n\to\infty} \frac{1-\left(1-\frac1n\right)^k}{2}= \frac{1-(1-0)^k}{2}=0. \)
Pro \(k<l\) vytkneme \(n^l\):
\(\displaystyle \lim_{n\to\infty} \frac{n^k-(n-1)^l}{n^k+n^l}= \lim_{n\to\infty} \frac{n^{k-l}-\left(1-\frac1n\right)^l}{n^{k-l}-1}= \frac{0-(1-0)^l}{0-1}=-1. \)
Výsledek
Limita je rovna \( \begin{cases} 1 & \text{ pro } k>l,\\ 0 & \text{ pro } k=l,\\ -1 & \text{ pro } k<l. \end{cases} \)
Varianta 2
\(\displaystyle \lim_{n\to\infty} \frac{(n+1)^k+(-n)^l}{(n-1)^k-n^l}\)
Řešení
Pro \(k>l\) vytkneme \(n^k\):
\(\displaystyle \lim_{n\to\infty} \frac{(n+1)^k+(-n)^l}{(n-1)^k-n^l}= \lim_{n\to\infty} \frac{\left(1+\frac1n\right)^k+(-1)^ln^{l-k}}{\left(1-\frac1n\right)^k-n^{l-k}}= \frac{(1+0)^k+(-1)^l\cdot 0}{(1-0)^k-0}=1 \)
Pro \(k=l\) nejprve vytkneme \(n^k\):
\(\displaystyle \lim_{n\to\infty} \frac{(n+1)^k+(-n)^k}{(n-1)^k-n^k}= \lim_{n\to\infty} \frac{\left(1+\frac1n\right)^k+(-1)^k}{\left(1-\frac1n\right)^k-1}=\\ \lim_{n\to\infty} \frac{\left(1+\frac1n\right)^k+(-1)^k}{\left(1-\frac1n-1\right) \left[\left(1-\frac1n\right)^{k-1}+\left(1-\frac1n\right)^{k-2}+\cdots+ \left(1-\frac1n\right)+1\right]} \)
Je-li \(k\) sudé, dostaneme \(\displaystyle\lim_{n\to\infty} -n\cdot\frac{2}{k}=-\infty\).
Je-li \(k\) liché rozložíme podobně i čitatel \(\left(1+\frac1n\right)^k-1\) a dostaneme
\(\displaystyle \lim_{n\to\infty} \frac{ \left(1+\frac1n-1\right) \left[\left(1+\frac1n\right)^{k-1}+\left(1+\frac1n\right)^{k-2}+\cdots+ \left(1+\frac1n\right)+1\right]}{\left(1-\frac1n-1\right) \left[\left(1-\frac1n\right)^{k-1}+\left(1-\frac1n\right)^{k-2}+\cdots+ \left(1-\frac1n\right)+1\right]} = \lim_{n\to\infty} \frac{-n}{n}\cdot\frac{k}{k}=-1 \)
Pro \(k<l\) vytkneme \(n^l\):
\(\displaystyle \lim_{n\to\infty} \frac{(n+1)^k+(-n)^l}{(n-1)^k-n^l}= \lim_{n\to\infty} \frac{\left(1+\frac1n\right)^kn^{k-l}+(-1)^l}{\left(1-\frac1n\right)^kn^{k-l}-1}= \frac{(1+0)^k\cdot 0 +(-1)^l}{(1-0)^k\cdot 0-1}=(-1)^{l+1} \)
Výsledek
Limita je rovna \( \begin{cases} 1 & \text{ pro } k>l,\\ -\infty & \text{ pro } k=l \text{ sudé},\\ -1 & \text{ pro } k=l \text{ liché},\\ (-1)^{l+1} & \text{ pro } k<l. \end{cases} \)
