$f(x) \leq g(x), \forall x\in (a,\infty)$, prove that $\lim_{x\rightarrow\infty} f\leq$ lim$_{x\rightarrow\infty} g$ (Given both these limits exist)

My progress: Let $\lim_{x\rightarrow\infty} f=L$, and $\lim_{x\rightarrow\infty} g=M$.

$\therefore$ Given $\varepsilon>0$, $\exists k_1>a$ such that $|f(x)-L|<\varepsilon, \forall x>k_1$ and $\exists k_2>a$ such that $|g(x)-M|<\varepsilon, \forall x>k_2$

Let $k=\max\{ k_1,k_2\} \Rightarrow \forall x>k, |f(x)-L|<\varepsilon,|g(x)-M|<\varepsilon$

I need to show that $L\leq M$. I don't know how to go from here. Please help!

  • 2
    $\begingroup$ Are you allowed to use the sum theorem for limits? That is $\lim(f+g)=\lim(f)+\lim(g)$? If so then all you need to know is that if $f(x)\geq0$ $\forall$ $x$ then $\lim(f)\geq0$ which is easy to prove by contradiction. $\endgroup$ May 5, 2015 at 20:11
  • 3
    $\begingroup$ Note that writing \in with no space rather than \ in with a space (which is what I found here) gives you $\forall x\in(a,\infty)$ rather than $\forall x\ in(a,\infty)$, and I edited accordingly. Also, writing \max with a backslash not only de-italicizes it but provides proper spacing in things like $a\max b$ and affects formatting of subscripts in a "displayed" setting thus: $\displaystyle a\max_{c\in D}b$. And \lim_{x\to\infty} (with a backslash) appears as $\lim_{x\to\infty}$ in an inline setting and $\displaystyle\lim_{x\to\infty}$ in a displayed setting. ${}\qquad{}$ $\endgroup$ May 5, 2015 at 20:11
  • $\begingroup$ Assume $L>M$ and choose $\epsilon=\frac{L-M}{2}>0$. Show that this is impossible. $\endgroup$
    – pascalhein
    May 5, 2015 at 20:14
  • $\begingroup$ @MichaelHardy thanks a lot for the formatting tips! :) $\endgroup$
    – Diya
    May 5, 2015 at 20:17
  • 1
    $\begingroup$ @Diya yes, that is correct. $\endgroup$
    – pascalhein
    May 5, 2015 at 20:25

3 Answers 3


Suppose $f(x)\geq0$ $\forall$ $x$. Suppose $\lim\limits_{x\to\infty}f(x)=L<0$. Let $\epsilon=|L|/2$. Then $\exists$ $N$ such that $x>N$ $\Rightarrow$ $|f(x)-L|<|L|/2$. This is equivalent to $$L/2 < f(x)-L < -L/2$$ Thus $x>N$ $\Rightarrow$ $f(x) < L/2 < 0$. Contradiction. Now apply this to $g(x)-f(x)$.


Try with $\epsilon = |M - L| / 2$.


It is much simpler to argue via contradiction. I use your notation of $L, M$ and on the contrary assume that $L > M$. This means that $L - M > 0$ and hence we can choose a positive number $\epsilon = (L - M)/2$.

Since $\lim\limits_{x \to \infty}f(x) = L$ it follows that there is a number $k_{1} > 0$ such that $$L - \epsilon < f(x) < L + \epsilon$$ whenever $x > k_{1}$. Thus we have $$f(x) > \frac{L + M}{2}\tag{1}$$ whenever $x > k_{1}$.

Again $\lim\limits_{x \to \infty}g(x) = M$ implies the existence of a number $k_{2} > 0$ such that $$M - \epsilon < g(x) < M + \epsilon$$ whenever $x > k_{2}$. And then we have $$g(x) < \frac{L + M}{2}\tag{2}$$ whenever $x > k_{2}$.

If we take $k$ as the maximum of $k_{1}, k_{2}, a$ and keep $x > k$ then we have both $x > k_{1}$ and $x > k_{2}$ and hence both inequalities $(1)$ and $(2)$ are satisfied simultaneously when $x > k$. We then have $$g(x) < \frac{L + M}{2} < f(x)$$ for all $x > k \geq a$. But this contradicts the given assumption that $f(x) \leq g(x)$ for all $x \geq a$.

How does one arrive at such a proof? The idea is simple. If $\lim\limits_{x \to \infty} f(x) = L$ and $\lim\limits_{x \to \infty} g(x) = M$ then $f(x)$ is near $L$ and $g(x)$ is near $M$ for all sufficiently large values of $x$. If $L > M$ then there will come a situation when $f(x)$ is so near to $L$ as to exceed $M$ and $g(x)$ is so near to $M$ as to be exceeded by $L$. The situation on number line will turn out to be like

-------|--------M--------|-------$\dfrac{L + M}{2}$-------||-------L--------||-------

Here single "|" represents the zone of values of $g(x)$ near $M$ and double "||" represents the zone of values of $f(x)$ near $L$. And it is clear that in this case $g(x) < f(x)$. In order that the zone of $g(x)$ remains separate from the zone of $f(x)$, it is necessary to find the mid point $(L + M)/2$ of $L, M$ on number line and ensure that the zone of $g(x)$ is left of this midpoint and zone of $f(x)$ is to the right of this midpoint.

In order to achieve this it is necessary to keep the distance between $g(x)$ and $M$ to be less than the distance of $M$ from this midpoint value $(L + M)/2$ and the required distance is $(L - M)/2$. Same goes for distance between $f(x)$ and $L$. This is how we arrive at $\epsilon = (L - M)/2$ and formalize (and sometimes intimidate a beginner learning calculus) a simple argument about inequalities.

  • $\begingroup$ +1 specially for the "behind the scenes" part. My life as a begginer in real analysis would have been so much easier if an explanation like this were given the early proofs. $\endgroup$
    – user561334
    Sep 1, 2018 at 5:53
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    $\begingroup$ @dude3221: I am glad that you liked that explanation. This is something which the teachers / instructors should provide in any introductory real analysis class. However the reality in universities may be different and I can't comment on that as I never studied real analysis in a formal classroom setting. $\endgroup$
    – Paramanand Singh
    Sep 1, 2018 at 7:04

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