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Suppose that we have function $f(x)$ ,which is defined as follows: $f(x):= \begin{cases} a+bx &\text{, if } x>2\\ 3 &\text{, if } x=2\\ b-ax^2 &\text{, if } x<2\end{cases}$ (here $a,b$ are some constants).

We should find $a,b$ so that limit for $x\to 2$ of $f(x)$ exists and equals $3$. I think that, because limit at point $2$ exist, it means that left and right limits are equal, so after we evaluate limits on left and right side, we will get $a+2b=b-4a\mbox{ or }b=-5a.$ Also because the limit is equal to $3$ at point $2$, it means that left or right limit is also equal to $3$, so $a+2b=3$ put one in another, I have got that $a=-1/3$ and $b=5/3$. Am i correct?

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    yes @Paul it is exactly what i wanted2011-12-23

1 Answers 1

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Yes, your reasoning is correct. We have $\lim_{x\rightarrow 2^-}f(x)=\lim_{x\rightarrow 2^-}(b-ax^2)=b-4a,$ $\lim_{x\rightarrow 2^+}f(x)=\lim_{x\rightarrow 2^+}(a+bx)=a+2b.$ Therefore, if $\displaystyle\lim_{x\rightarrow 2}f(x)=3$, we have $\lim_{x\rightarrow 2^-}f(x)=\lim_{x\rightarrow 2^+}f(x)=\lim_{x\rightarrow 2}f(x)=3,$ which implies that $b-4a=a+2b=3.$ Solving these linear equations, we get $a=-1/3$ and $b=5/3$.

One more thing, $\displaystyle\lim_{x\rightarrow 2}f(x)=3=f(2)$ means that $f$ is continuous at $2$.