「 Is the function \( f(x) \) continuous at \( x=2 \) for \[ f(x)=\left\{\begin{array}{ll}\sqrt{2+x} & \text { for } x>2 \\ 7 & \text { for } x=2 \\ (x+1)^{2}-5 & \text { for } x<2\end{array}\right. \] One of your students submitted the following solution: Solution: \[ \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} \sqrt{2+x}=\sqrt{2+2}=4 \] Function \( f(x) \) is continuous at \( a \) if \( \lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x) \). For \( a=2 \) we have \[ f(x)=\lim _{x \rightarrow 2^{-}}(x+1)^{2}-5=(2+1)^{2}-5=4 \] Therefore, right-hand and left-hand limits equal and the function is continuous. \( ] \) Please find a mistake in student's solution and provide your correct version.
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