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प्रश्न
f(x) = `{{:((1 - cos 2x)/x^2",", "if" x ≠ 0),(5",", "if" x = 0):}` at x = 0
उत्तर
We have, `{{:((1 - cos 2x)/x^2",", "if" x ≠ 0),(5",", "if" x = 0):}` at x = 0
At x = 0
L.H.L. = `lim_(x -> 0^-) (1 - cos 2x)/x^2`
= `lim_("h" -> 0) (1 - cos2(0 - "h"))/(0 - "h")^2`
= `lim_("h" -> 0) (1 - cos 2"h")/"h"^2`
= `lim_("h" -> 0) (2 sin^2"h")/"h"^2` = 2
R.H.L. = `lim_("h" -> 0^+) (1 - cos 2x)/x^2`
= `lim("h" -> 0) (1 - cos2(0 + "h"))/(0 + "h")^2`
= `lim_("h" -> 0) (2sin^2"h")/"h"^2` = 2
Also f(0) = 5 ... (Given)
Since, L.H.L. = R.H.L. ≠ f(0)
Hence, f(x) is discontinuous at x = 0.
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