Prove that for every $c>0$ and for every polynomial $p(x) \in \mathbb{R}[x]$ the limit $\lim\limits_{x \to \infty}{\frac{p(x)}{e^{cx}}}$ exists and is eqaual to $0$.

Use the L'Hospital's Rule and mathematical induction by the degree $k$ of the polynomial $p(x)$.


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  • $\begingroup$ What happens if you follow the given hint? $\endgroup$ – Hagen von Eitzen Jan 6 '16 at 17:30
  • $\begingroup$ Tried to start with the indcution for k=0 but then I get something like: $\lim\limits_{x\to \infty}{\frac{a_0}{e^{cx}}}$ Does it even equals 0? $\endgroup$ – Monsieur Molly Jan 6 '16 at 17:41
  • $\begingroup$ @MonsieurMolly $e^x$ goes to $\infty$ when $x\to\infty$. (also, degree of a polynomial -- I've never seen the word "grade" for that (?)) $\endgroup$ – Clement C. Jan 6 '16 at 17:54

We know that the statement is true for the polynomial of degree zero (just a constant $b_0$) $$ \lim_{x\to\infty}\frac{b_0}{e^{cx}}=b_0\lim_{x\to\infty}\frac{1}{e^{cx}}=0\ . $$ Now, assume the property is true for a polynomial of degree $k-1$, $$ \lim_{x\to\infty}\frac{\sum_{m=0}^{k-1}b_m x^m}{e^{cx}}=0\ . $$ Then, for a polynomial of degree $k$ we have $$ \lim_{x\to\infty}\frac{\sum_{m=0}^{k}b_m x^m}{e^{cx}}=\lim_{x\to\infty}\frac{\sum_{m=0}^{k-1}b_m x^m+b_k x^k}{e^{cx}}=\lim_{x\to\infty}\left[\frac{\sum_{m=0}^{k-1}b_m x^m}{e^{cx}}+\frac{b_k x^k}{e^{cx}}\right]\ . $$ The first summand goes to zero by the induction hypothesis. For the second summand, just apply L'Hospital $k$ times. Differentiating $x^k$ $k$-times reduces the numerator to a constant, while the denominator remains an exponential. So again the limit is zero.

  • 3
    $\begingroup$ You could just apply L'Hopital to the polynomial and obtain a polynomial of degree k-1. There you can apply the induction hypothesis. $\endgroup$ – Paul K Jan 6 '16 at 17:52

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