# Prove the limit using mathematical induction and L'Hospital's rule. [closed]

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 $p(x)$.

## closed as off-topic by Chappers, Aloizio Macedo♦, Fabian, colormegone, 6005Jan 6 '16 at 21:20

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• What happens if you follow the given hint? – Hagen von Eitzen Jan 6 '16 at 17:30
• 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? – Monsieur Molly Jan 6 '16 at 17:41
• @MonsieurMolly $e^x$ goes to $\infty$ when $x\to\infty$. (also, degree of a polynomial -- I've never seen the word "grade" for that (?)) – 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.