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प्रश्न
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous :
f(x) `{:(=("e"^(5sinx) - "e"^(2x))/(5tanx - 3x)",", "for" x ≠ 0),(= 3/4",", "for" x = 0):}}` at x = 0
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उत्तर
f(0) = `3/4` ...(Given)
`lim_(x -> 0) "f"(x) = lim_(x -> 0) ("e"^(5sinx) - "e"^(2x))/(5tanx - 3x)`
= `lim_(x -> 0) (("e"^(5sinx) - 1) - ("e"^(2x) - 1))/(5tanx - 3x)`
= `lim_(x -> 0) [((("e"^(5sinx) - 1) - ("e"^(2x) - 1))/x)/((5tanx - 3x)/x)]` ...`[("Divde numerator and denomiantor by" x),(because x -> 0"," therefore x ≠ 0)]`
= `(lim_(x -> 0) (("e"^(5sinx) - 1)/x - ("e"^(2x) - 1)/x))/(lim_(x -> 0) ((5tanx)/x - 3))`
= `(lim_(x -> 0)("e"^(5sinx - 1)/(5sinx) (5sinx)/x) - lim_(x -> 0) (("e"^(2x) - 1)/(2x) xx 2))/(lim_(x -> 0) (5tanx)/x - lim_(x -> 0) 3)`
= `(5 lim_(x -> 0) (("e"^(5sinx) - 1)/(5sinx))* lim_(x -> 0) (sinx/x) - 2 lim_(x -> 0) ("e"^(2x) - 1)/(2x))/(5 lim_(x -> 0) tanx/x - lim_(x -> 0) (3))`
= `(5(1)(1) - 2(1))/(5(1) - 3)` ...`[(because x -> 0"," 2x -> 0"," sinx -> 0"," 5sin x -> 0 and),(lim_(x -> 0) ("e"^(x - 1)/x) = 1 "," lim_(x -> 0) sinx/x = 1)]`
= `3/2`
∴ `lim_(x -> 0) "f"(x) ≠ "f"(0)`
∴ f(x) is continuous at x = 0.
∴ f(x) has a removable discontinuity at x = 0
This discontinuity can be removed by redefining f(0) = `3/2`.
∴ f(x) can be redefined as
f(x) `{:(=("e"^(5sinx) - "e"^(2x))/(5tanx - 3x)",", x ≠ 0),(= 3/2",", x = 0):}`
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