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Given that: $$\lim_{x\to0} \frac{f(x)}{x}=0$$ How can I prove that: $$\lim_{x\to0} f(x)=0$$ ?

Would the L'Hospital's Rule be applicable?

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    $\begingroup$ Given that the first limit exists, I think that $\lim_{x\to0}f(x)=0$ must hold. $\endgroup$ – Akiva Weinberger Apr 13 '15 at 8:38
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    $\begingroup$ thing which is given , we can write limit of [f(x)-0]/[x-0]=0, hence given thing is derivative of f(x) at 0 i.e. f'(0)=0 $\endgroup$ – Murtuza Vadharia Apr 13 '15 at 8:40
  • $\begingroup$ L'Hospital's rule applies to calculating derivatives, but you have no assumptions about $f(x)$ being differentiable.. $\endgroup$ – CiaPan Apr 14 '15 at 8:53
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Hint :

$f(x)=\frac{f(x)}{x}{x}$ and product rule.

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    $\begingroup$ Clearly the best method in this case. $\endgroup$ – GEdgar Apr 13 '15 at 12:41
  • $\begingroup$ Product rule is a rule of derivative calculations, but the problem assumes nothing about differentiability.... $\endgroup$ – CiaPan Apr 14 '15 at 8:51
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    $\begingroup$ @CiaPan By "product rule" I mean product rule for limits $\endgroup$ – John Apr 14 '15 at 10:12
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If by contradiction $f(x)\to a\neq0,\;\infty$, your limit $\lim f(x)/x$ would be $\infty$. How can you obtain a contradiction if $\lim f(x)$ doesn't exist?

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    $\begingroup$ ....How indeed? $\endgroup$ – Timbuc Apr 13 '15 at 9:38
  • $\begingroup$ It actually suffices to consider what happens if the limit doesn't tend to 0. You can construct a sequence $\{x_n\}$ tending to $0$ such that $f(x_n)<\epsilon$ where $\epsilon$ is fixed. $\endgroup$ – ktoi Apr 13 '15 at 10:14
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By definition, if $\lim_{x \to 0} f(x)/x = 0$ then for all $\epsilon>0$ there exist small enough $x$ such that

$$-\epsilon<f(x)/x<\epsilon$$ therefore, since $x\ne0$ $$-\epsilon\cdot x<f(x)<\epsilon\cdot x$$ and when $x\in (0, 1)\cup(-1, 0)$ we get $$-\epsilon<f(x)<\epsilon$$ which implies that $\lim_{x \to 0} f(x) = 0$

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Because the $\lim_{x\to0} \frac{f(x)}{x}=0$, this means that $f(x)$ tends to $0$ faster than $x$, or that $|f(x)|\le x$.

Therefore:

$$0\le|f(x)|\le x $$

Using the squeeze theorem, and because $\lim_{x\to0} x=0$, $\lim_{x\to0} f(x){x}=0$.

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    $\begingroup$ Although there may exist a point of view in which your first sentence is right, I believe that it is not at all a rigorous thing to say (i.e. if I get an homework with those words, I won't never give full mark). This said, in order to deduce $|f(x)<x$ there must be some work required (perhaps using the formal definition of limit applied to the hypothesis). $\endgroup$ – rafforaffo Apr 13 '15 at 11:30
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    $\begingroup$ To expand on rafforaffo's comments: there is nothing wrong with saying $|f(x)| < x$ for $x$ sufficiently close to zero (well, if you justify the statement formally, at any rate), but it may not hold in general. For example, let $f(x) = x^2$. $\endgroup$ – Strants Apr 13 '15 at 18:15
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By the definition of a limit, f (x) / x < eps for every eps if x is small enough. Take eps = 1, so f (x) / x < 1 if x < eps1, or f (x) < x if x < eps1. Take the definition of the limit again; f (x) < eps if you take x < min (eps, eps1).

Obviously you don't need that the limit of f (x) / x is 0. If the limit is c, then f (x) / x < c+1 for small x, so f (x) < x * (c + 1) for small x, and f (x) < eps if x < eps / (c + 1).

Obviously you'll have to add a few absolute values in the argument.

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This is not a fancy proof. It is a grinding one from the epsilon-delta definition of limit suitable for an introductory calculus class.

The core of the idea is that if $f(x)$ is bounded away from $0$ as $x$ goes to zero, then $f(x)/x$ is also bounded away from zero, as for small $x$, dividing by $x$ makes things further from zero.

And things bounded away from zero don't have a limit of zero.


Suppose the limit as $x$ goes to zero of $f(x)$ is $a$, and $a > 0$.

Let $\epsilon = a/2$. Then there is a $\delta$ such that for all $0 < x < \delta$, $a/2 < f(x) < 3a/3$.

Let $\delta_0$ be the minimum of $1/2$ and $\delta$.

(statement 1): Then for $0 < x < \delta_0$, $f(x)/x \geq f(x)*2 \geq 2a/2 = a$

If the limit of $f(x)/x$ as $x$ goes to zero is $0$, then let $\epsilon = a/2$.

(statement 2): Then there exists a $\delta_1 > 0$ such that for all $0 < x < \delta_1$, $-a/2 < f(x)/x < a/2$

Let $\delta_2$ be the least of $\delta_0$ and $\delta_1$.

For $x < \delta_2 <= \delta_1$, $f(x)/x < a/2$ by (2).

For $x < \delta_2 <= \delta_1$, $f(x)/x \geq a$ by (1).

Thus $a/2 > a$ or $1/2 > 1$, a contradiction.

A similar result holds of $a < 0$. Spotting the places where I implicitly assumed $a$ was positive may be interesting.

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