This is a question taken out of one of MIT's Single Variable Calculus Exams in 2010 (Obtained via OCW)

Problem : Find $f(0)$, given that $ f $ is continuous at $ x=0$ and $$\lim_{x \to 0} \frac{f(x)}{x} = 1$$

Solution: The official solution to this problem is quoted verbatim below :

Since $\lim_{x \to 0} \frac{f(x)}{x} = 1$, $f(x) \to 0$ as $x \to 0$, so $f(0)=0$

However what this solution has done (at least to me it seems this way), is simply, directly substitute $x=0$, into $\frac{f(x)}{x}$, and then solve for $f(0)$, but the way in which it is done, doesn't seem very rigorous (at least to me it doesn't).

This is essentially what the solution has done (at least it seems this way to me)

\begin{aligned} & \lim_{x \to 0}\ \frac{f(x)}{x} = 1 \\ & \implies \frac{f(0)}{0} = 1 & \text{(By direct substitution)} \\ & \implies f(0) = 0 \cdot 1 & \text{(Can we even do this?)}\\ & \implies f(0) = 0\\ \end{aligned}

But is what is being done in the second and third steps even correct? I ask this because we have a denominator of $0$ on one side of the equation, and technically having $0$ as the denominator in one of the terms of an equation, makes the equation undefined, am I correct in saying this?

Essentially my question boils down to the fact that division by $0$ is undefined, and I am asking whether algebraically manipulating an equation that has one term, with a denominator of $0$, is mathematically valid or not, and whether this is seen as a rigorous solution to the posed problem.

  • $\begingroup$ You cannot "plug in 0," what you write when doing so has a division by zero -- it is not mathematically well-defined. (Now, checking something else -- is $f$ assumed to be continuous?) $\endgroup$ – Clement C. May 12 '16 at 15:54
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    $\begingroup$ @Clement C, I thought as much, yes $f$ is assumed to be continuous $\endgroup$ – Perturbative May 12 '16 at 15:57
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    $\begingroup$ From the fact that $\lim_{x \to 0}\dfrac{f(x)}{x} = 1$ we can only conclude that $\lim_{x \to 0}f(x) = 0$. We can't conclude anything about $f(0)$ unless we have some more information. If $f$ is continuous at $x = 0$ then $f(0) = 0$. $\endgroup$ – Paramanand Singh May 12 '16 at 15:57

As some comments have pointed out, it is possible for $f(x)$ to have a discontinuity at $0$ and $f(0) \ne \lim_\limits{x\to 0} f(x)$

However, if $\lim_\limits{x\to a} \frac{f(x)}{g(x)} = M$ with $M$ bounded and $|M|>0$, then $\lim_\limits{x\to a} g(x) = 0 \iff \lim_\limits{x\to a} f(x) = 0$

If you want to be rigorous then you really should to recall the definition of limits.

$\forall \epsilon>0,\exists \delta>0$ such that $|x|<\delta \implies |\frac{f(x)}{x} - 1|<\epsilon$

$|f(x) - x|<\epsilon |x|$

  • $\begingroup$ Is this quoted section of your answer a theorem? "However, if $\lim_\limits{x\to a} \frac{f(x)}{g(x)} = M$ with $M$ bounded and $|M|>0$, then $\lim_\limits{x\to a} g(x) = 0 \iff \lim_\limits{x\to a} f(x) = 0$", I have never seen this before, and would explain the difficulty I'm facing in attempting to solve this problem. $\endgroup$ – Perturbative May 12 '16 at 16:28
  • $\begingroup$ You can make it one. It shouldn't be too hard to prove. $\endgroup$ – Doug M May 12 '16 at 17:09

By the product rule for limits, as these limits exist,

$$\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac{f(x)}{x}\cdot\lim_{x \to 0} x =1\cdot0.$$

But it is impossible to conclude anything about $f(0)$.

  • $\begingroup$ How does this show $f(0)=0$? All you've shown is $\lim_{x\to 0} f(x)=0$ $\endgroup$ – user223391 May 12 '16 at 16:24
  • $\begingroup$ @ZacharySelk: you are right. Actually, no limit gives information on $f(0)$, unless continuity at $0$ is stated. $\endgroup$ – Yves Daoust May 12 '16 at 16:27

If I found the correct exam on OCW, there's some important information from the question that was left out. I think this is question 6 from this exam: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/exam-1/session-22-materials-for-exam-1/MIT18_01SCF10_exam1.pdf

The question is actually: "Suppose that $f$ satisfies the equation $f(x+y) = f(x) +f(y) + x^2 y +xy^2$ for all real numbers $x$ and $y$. Suppose further that $$ \lim_{x \rightarrow 0} \frac{f(x)}{x} = 1. $$ Find $f(0)$."

So you can get the answer directly by plugging in $x=y,y=0$ into the given equation $f(0+0)=f(0)+f(0) +0 +0$ which gives $f(0)=2f(0)$ or $f(0)=0$.

  • $\begingroup$ that is the correct question, however I omitted the rest of the question, as the solution only uses the limit to evaluate $ f (0) $, which is the reason why I asked this question as it was not clear to me, how they arrived at the given solution using the limit. I wanted this question to be aimed directly at asking how the limit could be used to evaluate $ f (0) $. For the purposes of Stack Exchange, I believed the other information would be irrelevant to the question I was asking. $\endgroup$ – Perturbative May 12 '16 at 17:04
  • $\begingroup$ Thanks Peter, I took my assertion regarding MIT in an earlier comment off now. $\endgroup$ – imranfat May 12 '16 at 17:10
  • $\begingroup$ @hardmath I've added the necessary assumption of continuity to the body of the question. Thanks for pointing that out. $\endgroup$ – Perturbative May 12 '16 at 17:16

Here is a very simple counter example. Consider $f(x)=\frac{sinx^2}{x}$ It is easy to see that $f(0)$ does NOT exist. As a limit, sure it does (it is zero) but $x=0$ is obviously not in the domain. However if you use this $f(x)$ in your limit we arrive at the expression $\frac{sinx^2}{x^2}$ and as a limit when $x$ approaches zero, this limit is $1$


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