If f(x)={|x+2|/tan^-1(x+2),x≠-2;2,x=-2 then discuss the continuity of f(x) at x=-2

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f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{| x + 2 |}{{tan}^{- 1} (x + 2)}, & x \neq - 2 2, & x = - 2 \end{matrix}

f (x)

f (x) = \{\begin{cases} \frac{|x + 2|}{\tan^{- 1} (x + 2)} & , x \neq - 2 \\ 2 & , x = - 2 \end{cases}

f (x) = ⎡ ⎢ ⎣ \begin{matrix} 1 - x & , & (0 \leq x \leq 1) x + 2 & , & (1 < x < 2) 4 - x & , & (2 \leq x \leq 4) \end{matrix} ⎤ ⎥ ⎦

y = f (x)

0 \leq x \leq 4

f (x) = {\begin{matrix} \frac{| x + 2 |}{{tan}^{- 1} (x + 2)} & for x \neq - 2 2 & for x = - 2 \end{matrix}

f (x)

x = 0

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x - | x |}{x}; x \neq 0 x = 2; x < 0 \end{matrix}

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