If f(x)={x2x≠25x=2f(x) = \begin{cases} x^2 & x \neq 2 \\ 5 & x = 2 \end{cases}f(x)={x25x=2x=2, then limx→2f(x)\lim_{x \to 2} f(x)limx→2f(x) equals:
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Does not exist
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