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?