# Continuous function, not sure what to do here…

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...

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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 $$f(1)\times g(1)=\lim_{x\to 1^-}f(x)g(x)=\lim_{x\to 1^-}f(x)\times \lim_{x\to 1^-}g(x)\to\lim_{x\to 1^-}g(x)=f(1)\times g(1)\\=2\times\frac{5}{2}=5$$

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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 $\lim_ {x\to 1+} g(x) = 5/2$.

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