हिंदी

The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous : f(x) =e5sinx-e2x5tanx-3x, for x≠0=34, for x=0} at x = 0

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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Continuity - EXERCISE 8.1 [पृष्ठ १७३]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
EXERCISE 8.1 | Q 9) (i) | पृष्ठ १७३

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