Prove $\lim_\limits{x\to\infty}\dfrac{P_k(x)}{P_{k+1}(x)}=0$ by limits.
$P_k(x)$ is defined as a polynomial of degree $k$.
Asked
Active
Viewed 72 times
0
-
See Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials – Bart Michels Apr 17 '15 at 07:46
-
possible duplicate of Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials – Peter Taylor Apr 17 '15 at 09:24
2 Answers
4
Methodology Proof
Divide the numerator and denominator by $x^k$. Then the numerator has a constant term and a finite number of of terms of the form $$ \frac{a_j}{x^j} \to 0 \quad \text{as } \quad x \to \infty$$ while the denominator has a linear term which goes to $\infty$.
davidlowryduda
- 91,687
2
$$p_k=a_1x^k+a_2x^{k-1}+...\\p_{k+1}=b_1x^{k+1}+b_2x^{k}+...\\ \lim_{x \rightarrow \infty} \frac{p_{k}}{p_{k+1}}=\lim_{x \rightarrow \infty} \frac{a_1x^k+a_2x^{k-1}+...}{b_1x^{k+1}+b_2x^{k}+...}=\\\lim_{x \rightarrow \infty} \frac{x^K(a_1+a_2\frac{1}{x}+a_3\frac{1}{x^2}...)}{x^{k+1}(b_1+b_2\frac{1}{x}+b_3\frac{1}{x^2}..)}=\\ \lim_{x \rightarrow \infty} \frac{x^k(a_1)}{x^{k+1}(b_1)}=\\ \lim_{x \rightarrow \infty} \frac{(a_1)}{x^{1}(b_1)}\rightarrow 0 \\$$
Khosrotash
- 24,922