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Let $f,g:\mathbb{R}\to\mathbb{R}$ denote the functions

$$f(x)=\begin{cases}-x+3,&x\leq 1,\\-ax^2,&\text{otherwise}.\end{cases}$$ $$g(x)=\begin{cases}\frac{x^3+x^2-x-1}{x-1},&x\neq 1,\\-b,&\text{otherwise}.\end{cases}$$

Determine parameters $a,b\in\mathbb{R}$ such that $f$ and $g$ are continuous.

I was thinking about comparing limits for both sides, that is finding $a$ via

$$\lim\limits_{x\to 1}-x+3=\lim\limits_{x\to 1}-ax^2$$

and $b=-4$ with the same approach. Actually it looked obvious to me that $a=-2$, however my thoughts don't comply with the solution of WolframAlpha. What am I doing wrong here?

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    $\begingroup$ Believe in you, not in Wolfram. As $\lim_{x\to 1} 3-x = 2 = \lim_{x\to 1} 2x^2$, you are right. By the way, $a = -2$, then. $\endgroup$
    – martini
    Nov 10, 2012 at 13:32
  • $\begingroup$ Oh typo. Fixed it now - may I ask you whether you can confirm my second result $b=4$ which I did determine the same way as $a$? $\endgroup$ Nov 10, 2012 at 13:35
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    $\begingroup$ As for $a$, up to a sign, as $g(x) = (x+1)^2$ for $x \ne 1$, so the limit $\lim_{x\to 1, x\ne 1} g(x) = (1+1)^2 = 4 \stackrel != -b$, so $b=-4$. $\endgroup$
    – martini
    Nov 10, 2012 at 13:37
  • $\begingroup$ @martini: Oh again a typo... but nevertheless thanks. $\endgroup$ Nov 10, 2012 at 13:38

1 Answer 1

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Yes, $a=-2$ and $b=-4$.

Of course, $f$ and $g$ are continuous in the given point if the left and right limits there exist and coincide, and the plot of Wolfram also confirms this (though the text itself 'Discontinuity at $x=1$' is really misleading).

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