Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is as follows:

Let $f(x) = \begin{cases} x, & \mbox{if } x<1 \\ x^2+1, & \mbox{if } x\ge 1 \end{cases}$

Let $g$ be a function such that $fg$ is continuous at $1$, and $\displaystyle g(1)=\frac{5}{2}$.

What is $\displaystyle \lim_ {x\to 1-} g(x)$?

I have absolutely no idea what this is talking about...

share|cite|improve this question
It is not possible to find $\lim_{x\to -1}g(x)$ with the information given. Maybe you mean $\lim_{x\to 1}g(x)$? – Emanuele Paolini Feb 4 '13 at 6:33
Probably. I might just be reading it wrong. – samm_hall Feb 4 '13 at 6:34
@user60899, you should not post the same question twice. When you edit a post, it comes back up to the front page. I have merged your duplicate question into this one. – Zev Chonoles Feb 4 '13 at 6:59
This is really hurting my head... – samm_hall Feb 4 '13 at 7:03
@samm_hall Please accept an answer if it satisfies you. – Git Gud Feb 4 '13 at 7:36

Hint: For $x<1$, $~~f(x)$ is a continuous function and $$\lim_{x\to 1^-}f(x)=1$$ Also $$\lim_{x\to 1^+}f(x)=1^2+1=2$$ Now we have $$f(1)\times g(1)=\lim_{x\to 1^-}f(x)g(x)=\lim_{x\to 1^-}f(x)\times \lim_{x\to 1^-}g(x)\to\lim_{x\to 1^-}g(x)=f(1)\times g(1)\\=2\times\frac{5}{2}=5 $$

share|cite|improve this answer
So the limit is f(1) x g(1)? So it's just...1? – samm_hall Feb 4 '13 at 6:59
I just need to figure out what the limit of g(x) is as x approaches 1 from the left... – samm_hall Feb 4 '13 at 7:09
@samm_hall Following up on Babak's answer: the limit you're looking for doesn't exist. The lateral limits, however, are described in the answer. – Git Gud Feb 4 '13 at 7:22
@samm_hall: Limit of $g$ at $1^-$ is $2\times 2.5=5$ and the limit of it when $x\to 1^+$ is $\frac{2\times 2.5}{2}=2.5$. – Babak S. Feb 4 '13 at 7:23
Nice work and follow through (with comments/interaction with the OP! +1 – amWhy Feb 5 '13 at 2:27

I've edited the problem statement, since I'm betting that the problem asks to find $\lim_ {x\to 1-} g(x)$.

Note that $fg(1) = 5$. Because $fg$ is continuous at $1$, we must therefore have $\lim_{g\to1-} fg(x) = 5$. But $\lim_ {x\to 1-} f(x) = \lim_ {x\to 1-} x = 1$, and so $$ \lim_ {x\to 1-} g(x) = \lim_ {x\to 1-} \frac{fg(x)}{f(x)} = \frac{\lim_ {x\to 1-}fg(x)}{\lim_ {x\to 1-}f(x)} = \frac51 = 5. $$ (Similarly one can show that $\lim_ {x\to 1+} g(x) = 5/2$.

share|cite|improve this answer
You did it via another way +1. You wrote somewhere $g\to 1^-$ – Babak S. Feb 4 '13 at 8:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.