Let $f(x)$ be a continuous function in the real number set, such that

$$f(x)=\begin{cases} \pi, \,\, \text{if} \,\, x> 1\\ g(x), \,\,\text{if}\,\,x≤ 1\end{cases}$$

Which of these statements are always correct?

1. $$\lim_{x\to 1} g(x)=\pi$$

2. $$g(1)=\pi$$

3. $$\lim_{x\to 1^-}\frac{f(x)}{g(x)}=1$$

My attempts.

Maybe I can not see the right connection about the continuity between $f(x)$ and $g(x)$ . That is my first problem.

We have $$f(1)=\lim_{x\to 1^+} f(x)=\pi=\lim_{x\to\ 1^-}f(x)=g(1)\implies \pi=g(1) $$

I think this doesn't imply,

$\lim_{x\to 1} g(x)=\pi$.

I want to say that about $3$, that is not always correct. Because, it can be

$$\lim_{x\to 1^-}g(x)=0$$

But, $$g(1)=\pi$$

So, my answer is $2$. It is possible that, I am completely wrong. I am not sure, what is going on here, exactly.


1 Answer 1


1. Since $g$ is only referenced in the definition of $f$ for $x \leq 1$, we don't know anything about $g$ for $x > 1$. In particular, we don't know $$ \lim_{x \to 1^+} g(x), $$ so we cannot guarantee that the two-sided limit exists, let alone that its value is $\pi$. For an explicit counterexample, use any function $g$ with a jump discontinuity at $x=1$ but continuous from the left: $$ g(1) = \lim_{x \to 1^-} g(x) = \pi $$ but $$ \lim_{x \to 1^+} g(x) $$ either doesn't converge or has limit $L \neq \pi$.

2. Your calculation is correct, and continuity of $f$ forces $g(1) = \pi$.

3. For $x < 1$, we have $f(x) = g(x)$, so $$ \lim_{x \to 1^-} \frac{f(x)}{g(x)} = \lim_{x \to 1^-} 1 = 1. $$

  • 4
    $\begingroup$ An important subtlety for (3): the function $\frac{f(x)}{g(x)}$ is $1$ whenever $g(x)$ isn't zero, so it is important that $g(x)$ is nonzero in a relevant neighborhood to the left of $1.$ However, because $\lim_{x \to 1^-} g(x) = \pi,$ we're guaranteed that there is some $\delta$ such that $\delta < x < 1$ implies that $|g(x) - \pi| < \pi \Rightarrow 0 < g(x) < 2\pi$ implying that $g(x)$ must be nonzero sufficiently close to $x = 1.$ $\endgroup$ Jun 15 at 22:46
  • $\begingroup$ I couldn't understand $3$.. $\endgroup$
    – User
    Jun 15 at 23:43
  • 2
    $\begingroup$ @User On a bit more of a big-picture scale, because $g(x)$ is approaching $\pi$ as $x$ approaches $1$ from the left, if we look at values close enough to $x = 1$ (but still less than $1$) then $g(x)$ has to eventually not be zero anymore, because it has to get up to $\pi.$ My comment from before is just how we would say that with the more strict, formal definition of limits rather than a looser, more intuitive one. $\endgroup$ Jun 15 at 23:56
  • $\begingroup$ @StephenDonovan So am I right that ,actualy we need to show that $g(x)$ is nonzero? $\endgroup$
    – User
    Jun 16 at 0:24
  • 1
    $\begingroup$ @User We do, but it follows from properties you've already shown $g$ has $\endgroup$ Jun 16 at 0:32

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