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प्रश्न
f(x) = `{{:(|x|cos 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
उत्तर
We have, `{{:(|x|cos 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}`
At x = 0
L.H.L. = `lim_(x -> 0^-) |x| cos 1/x`
= `lim_("h" -> 0) |0 - "h"| cos 1/(0 - "h")`
= `lim_("h" -> 0) "h" cos 1/"h"`
= 0 × [an oscillating number between –1 and 1] = 0
R.H.L. = `lim_(x -> 0^+) |x| cos 1/x`
= `lim_("h" -> 0) |0 + "h"| cos 1/(0 + "h")`
= `lim_("h" -> 0) "h" cos 1/"h"`
= 0 × [an oscillating number between –1 and 1] = 0
Also f(0) = 0 ....(Given)
Thus, L.H.L. = R.H.L. = f(0)
So, f(x) is continuous at x = 0
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