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Question
f(x) = `{{:(3x + 5",", "if" x ≥ 2),(x^2",", "if" x < 2):}` at x = 2
Solution
We have, f(x) = `{{:(3x + 5",", "if" x ≥ 2),(x^2",", "if" x < 2):}` at x = 2.
At x = 2
R.H.L. = `lim_(x -> 2^+) (3x + 5)`
= `lim_("h" -> 0) [3(2 + "h") + 5]` = 11
And L.H.L. = `lim_(x -> 2^-) x^2`
= `lim_("h" -> 0) (2 - "h")^2` = 4
Since, L.H.L. ≠ R.H.L. at x = 2
So, f(x) is discontinuous at x = 2.
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