$\lim_{x\to \infty}\dfrac{x^n}{x+e^x} $ without using LHR

Question

Prove that for each n∈N. $\lim_{x\to \infty} \dfrac {x^n}{x+e^x}=0$ using basic $\varepsilon$ - $N$ method. I tried as We have $\left|\dfrac{x^n}{x+e^x}\right|<\varepsilon$. I take $x>1$ and it become $\left|\dfrac{1}{x+e^x}\right|<\varepsilon$. Using $e^x>x$ I got $\left|\dfrac{1}{2e^x}\right|<\varepsilon$. From that I got $N=\ln\left(\dfrac{1}{2\varepsilon}\right)$. But my problem is using this inequality I can choose a $\varepsilon$ such that $x$ can become $<1$. I want to prove this for every positive $x$ without using L hospital rule, Can you help me?

Answer 1

Based on the comments exchanged between me and OP, it seems that a detailed step by step answer is needed. I am trying to provide one here. Before we begin it is better to clarify the definition of limit of a function $f(x)$ as $x \to \infty$.

A function $f(x)$ is said to tend to a limit $L$ as $x$ tends to $\infty$ (written symbolically as $\lim_{x \to \infty}f(x) = L$ if for any given real number $\varepsilon > 0$ it is possible to find a positive real number $N$ such that $|f(x) - L| < \varepsilon$ whenever $x > N$.

So we have to choose an $N$ (corresponding to an $\varepsilon$) such that $x > N \Rightarrow |f(x) - L| < \varepsilon$. From the comments by OP, it seems that OP believes that the implication is in other direction i.e. $|f(x) - L| < \varepsilon \Rightarrow x > N$. But this is not the case. However there is another catch. In practice when we have to find an actual expression for $N$ in terms of $\varepsilon$ we try to proceed with the starting equation $|f(x) - L| < \varepsilon$ and try to see what limitation it implies on the values of $x$ and then hope to find that we get some limitation of the form $x > N$. Perhaps this procedure to find $N$ is probably the source of confusion for OP.

I will try to illustrate this point in reference to the current question. We have been provided with $f(x) = \dfrac{x^{n}}{x + e^{x}}$ and $L = 0$ and we have to verify that $L$ is in fact the limit of $f(x)$ as $ x \to \infty$ according to the definition given above. The verification has to be in form of finding an $N$ corresponding to an $\varepsilon$.

So we have to find an $N > 0$ such that $x > N$ implies $|f(x)| < \varepsilon$ i.e. $\left|\dfrac{x^{n}}{x + e^{x}}\right| < \varepsilon$. So we try to figure out how we can make $|x^{n}/(x + e^{x})|$ smaller than $\varepsilon$. Clearly we see that $|x^{n}/(x + e^{x})| = x^{n}/(x + e^{x}) < x^{n}/e^{x}$ and hence it is sufficient if we can make $x^{n}/e^{x}$ smaller than $\varepsilon$. Again we note that $$e^{x} = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n + 1}}{(n + 1)!} + \cdots$$ and hence $e^{x} > x^{n + 1}/(n + 1)!$ and therefore $x^{n}/e^{x} < (n + 1)!/x$. So it is sufficient to make $(n + 1)!/x$ smaller than $\varepsilon$ i.e. $(n + 1)!/x < \varepsilon$. Now this thing we can easily do if we make $x > (n + 1)!/\varepsilon$.

If we follow the logic written in bold lines above then we can see that taking $x > (n + 1)!/\varepsilon$ is sufficient for our initial goal of making $|x^{n}/(x + e^{x})| < \varepsilon$. It follows that if we put $N = (n + 1)!/\varepsilon$ then $x > N$ implies $|f(x)| < \varepsilon$. Thus $\lim_{x \to \infty}f(x) = 0$.

The problem here is technically simple but probably because of the symbolism of epsilons it looks like a tough exercise. One has to understand that what we need here is a little common sense and a bit knowledge of inequalities. I hope I have tried to provide a very simple answer (and sorry for the length).

Answer 2

Fix an $n\in\mathbb{N}$. We must show that for $x$ sufficiently large, $\frac{x^n}{x+e^x}<\epsilon$. If we make the denominator smaller, the expression gets larger. Using the Taylor series $e^x=1+x+\frac{1}{2}x^2+\ldots$, we have $$ \bigg|\frac{x^n}{x+e^x}\bigg| \leq\bigg|\frac{x^n}{x+1+x+\ldots+\frac{1}{n!}x^n+\frac{1}{(n+1)!}x^{n+1}}\bigg| \leq\bigg|\frac{x^n}{\frac{1}{(n+1)!}x^{n+1}}\bigg| =\frac{(n+1)!}{x} $$ Since $n$ is fixed, by choosing $x$ sufficiently large, we can make this expression as small as we like.