I want to prove this limit:$\lim_{x \to +\infty}(|P(x)|-e^x)=-\infty$. I can't use Hopital's rule, but only the fact that $a_{n}=(1+\frac{x}{n})^n$ and $b_{n}=(1+\frac{x}{n})^{n+1}$ converge to $e^x$.
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Hint $a_n$ is a polynomial of degree $n$ in $x$ – kingW3 Nov 26 '20 at 14:46
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What's wrong with Hopital's rule? That's how I learned to prove it – Nov 26 '20 at 17:00
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@expressjs123 nothing wrong, my professor said to find another way to prove it, without using Hopital! – rider Nov 26 '20 at 17:14
3 Answers
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Use the definition of a limit. $\lim_{x \to \infty} f(x) = -\infty$ means that given any arbitrarily large $N$, you can come up with an $M$ such that $x > M \to f(x) < -N$.
Let $d$ be the degree of $P(x)$. So $P(x) = \sum_{i = 0}^d a_i x^i$. Then $|P(x)| \leq \sum_{i = 0}^d |a_i| |x|^i$ and if $|x| > 1$ then $|P(x)| \leq |x|^d \sum_{i = 0}^d |a_i|$. Define $A = \sum_{i = 0}^d |a_i|$ as the sum of the absolute values of the coefficients of $P(x)$. We have $|x| \gt 1 \implies |P(x)| \geq A |x|^d$.
Given $N$ we want to find an $M$ that is big enough so that $x \gt M \implies A x^d - e^x \lt -N$ which is to say $e^x \gt Ax^d + N$.
At this point we need to use the definition of $e^x$, in particular the fact that $e^x \gt (1 + x/n)^n$. Take $n = d + 1$ and we have $x \gt 1 \implies e^x \gt (1 + x/(d + 1))^{d + 1} \gt x^{d + 1}/(d + 1)^{d + 1}$. (We could have taken $n$ much higher but that's all we need.)
So now we want to show that we can make $x^{d + 1}/(d + 1)^{d + 1} \gt Ax^d + N$. Multiply both sides by $(d + 1)^{d + 1}/x^d$ to get $x \gt (d + 1)^{d + 1}(A + N/x)$. Since we are assuming $x \gt 1$ it suffices to take $M = (d + 1)^{d + 1}(A + N)$.
Putting it together:
Given any $N$ let $M = \max \left\{ 1, (d + 1)^{d + 1}(A + N) \right\}$ where $d$ is the degree of $P(x)$ and $A$ is the sum of the absolute values of the coefficients in $P(x)$. Then $x \gt M \implies P(x) - e^x \lt A x^d - (1 + x/(d + 1))^{d + 1} \lt A x^d - x^{d + 1}/(d + 1)^{d + 1} = (A - x/(d + 1)^{d + 1}) x^d \lt (A - (A + N)(d + 1)^{d + 1}/(d + 1)^{d + 1})x^d = -N x^d \lt -N$
QED
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Regarding your first comment to your answer to Can a one-sided limit exist at a discontinuity point of a derivative? (answer now deleted, so I'm using this answer to tell you), see my answer to How discontinuous can a derivative be? – Dave L. Renfro Nov 27 '20 at 16:33
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Since $(1+\frac{x}{n})^n \to e^x $, we can show that $e^x =\sum_{n=0}^{\infty} \dfrac{x^n}{n!} $.
Then, for $x > 0$, $e^x \gt \dfrac{x^{n+1}}{(n+1)!} $ so $\dfrac{x^n}{e^x} \lt \dfrac{(n+1)!}{x} \to 0$.
marty cohen
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For any polynomial, $$\lim_{x\to\infty}\frac{P(x+1)}{P(x)}=1$$ so there is an $X$ such that
$$\forall x>X:\frac{P(x+1)}{P(x)}<2$$ and $P(x)$ is bounded by a geometric progression of constant ratio $2<e$.