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Could someone solve this $$f(x) = \begin{cases} x^3+a, & x<0 \\ a\sin\frac{\pi{x}}{3}+b, & 0\leq {x}<2 \\ 3, & x=2 \\ \log_2x^{b+1}, & x>2\\ \end{cases}$$ If $\lim_{x\to0}$ and $\lim_{x\to2}$ both exist, find the value of a and b.

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    $\begingroup$ I'm not the downvoter, but the the downvote is likely due to no attempts have been shown. You should include your attempt when you post the question $\endgroup$ Feb 16, 2020 at 5:10
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    $\begingroup$ MathJax hint: instead of commas to separate the expression from its range of applicability, use an ampersand (&). That will space them out nicely and align the ranges $\endgroup$ Feb 16, 2020 at 5:14

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We have $$\lim_{x\rightarrow 0^-}f(x)=a; \lim_{x\rightarrow 0^+}f(x)=b;$$ $$\lim_{x\rightarrow 2^-}f(x)=b+a\frac{\sqrt{3}}{2}; \lim_{x\rightarrow 2^+}f(x)=b+1$$

Because $\exists \lim_{x\rightarrow 0}f(x)$ and $\exists \lim_{x\rightarrow 2}f(x)$, then

$a=b$ and $b+1=b+a\frac{\sqrt{3}}{2}$ so $a=b=\frac{2\sqrt{3}}{3}$

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  • $\begingroup$ @Nigel: "I did it that way ..." If so, then you should have included your solution with your question, so that people could simply validate it for you. In general, you should show whatever work you have done, so that answerers have a better idea how to answer you, without wasting time (theirs or yours) explaining things you already know. It can be particularly irksome for an answerer to go through the trouble of solving a problem and formatting an elaborate, instructive answer only to have the asker comment "Yeah, I already did that." Be a considerate asker. Cheers! :) $\endgroup$
    – Blue
    Feb 16, 2020 at 6:40

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