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प्रश्न
Examine the following function for continuity at the indicated point.
f(x) = `{((x^2 - 4)/(x-2) "," if x ≠ 2),(0 "," if x = 2):}` at x = 2
उत्तर
f(x) = `(x^2 - 4)/(x-2)`, also given that f(2) = 0
`"L"[f(x)]_(x=2) = lim_(x->2^-) f(x)`
[∵ x = 2 – h, where h → 0, x → 2]
`= lim_(h->0)` f(2 - "h") ..[∵ x = 2]
`= lim_(h->0) ((2 - "h")^2 - 4)/((2-"h") - 2)`
`= lim_(h->0) (4 + "h"^2 - 4"h" - 4)/(2 - "h" - 2)`
`= lim_(h->0) ("h"^2 - 4"h")/(-"h")`
`= lim_(h->0) ("h"("h - 4"))/(- "h")`
`= lim_(h->0)` h - 4
`= lim_(h->0) (0 - 4)/(-1)` = 4
But `"L"[f(x)]_(x=2)` f(2) = 0
∴ `"L"[f(x)]_(x=2) ne` f(2)
∴ The given function is not continuous at x = 2.
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