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प्रश्न
If f(x) = `{(|x| + 1,x < 0), (0, x = 0),(|x| -1, x > 0):}`
For what value (s) of a does `lim_(x -> a)` f(x) exists?
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उत्तर
Given function:
f(x) = `{(|"x"| + 1,"x" < 0), (0, "x" = 0),(|"x"| -1, "x" > 0):}`
(i) at x = 0,
`lim_("x" → 0^-) f("x") = lim_("x" → 0^-) (1 - "x") = 1`
`lim_("x" → 0^+) f("x") = lim_("x" → 0^+) ("x" - 1) = -1`
`lim_("x" → 0^-) f("x") ≠ lim_("x" → 0^+) f("x")`
`lim_("x" → 1) f("x")` does not exist at x = 0.
(ii) When a < 0
`lim_("x" → "a"^-) f("x") = lim_("x" → "a"^-) (1 - "x") = 1 - "a"`
`lim_("x" → "a"^+) f("x") = lim_("x" → "a"^+) (1 - "x") = 1 - "a"`
∴ `lim_("x" → "a"^-) f("x") = lim_("x" → "a"^+) f("x")`
That is, `lim_("x" → "a") f("x") = 1 - "a"`
(iii) When a > 0
`lim_("x" → "a"^-) f("x") = lim_("x" → "a"^-) ("x" - 1) = "a" - 1`
`lim_("x" → "a"^+) f("x") = lim_("x" → "a"^+) ("x" - 1) = "a" - 1`
∴ `lim_("x" → "a"^-) f("x") = lim_("x" → "a"^+) f("x")`
Hence, `lim_("x" → "a") f("x") = "a" - 1`
Thus,
When, a < 0, `lim_("x" → "a") f("x") = 1 - "a"`
When, a > 0 `lim_("x" → "a") f("x") = "a" - 1`
Hence, there exists `lim_("x" → "a")` f(x) for all a, a ≠ 0.
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