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प्रश्न
f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
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उत्तर
We have, f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
At x = 0
L.H.L. = `lim_(x -> 0^-) ("e"^(1/x))/(1 + "e"^(1/x))`
= `lim_("h" -> 0) ("e"^(1/(0 - "h")))/(1 + "e"^(1/(0 - "h"))`
= `lim_("h" -> 0) ("e"^(1/"h"))/(1 + "e"^(- 1/"h"))`
= `("e"^(- oo))/(1 + "e"^(- oo))`
= `0/(1 + 0)`
= 0
R.H.L. = `lim_(x -> 0^+) ("e"^(1/x))/(1 + "e"^(1/x))`
= `lim_("h" -> 0) ("e"^(1/(0 + "h")))/(1 + "e"^(1/(0 + "h"))`
= `lim_("h" -> 0) ("e"^(1/"h"))/(1 + "e"^(1/"h"))`
= `lim_("h" -> 0) 1/("e"^(-1/"h") + 1)`
= `1/("e"^(-oo) + 1)`
= `1/(0 + 1)`
= 1
Thus, L.H.L. ≠ R.H.L. at x = 0.
So, f(x) is discontinuous at x = 0.
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