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Evaluate the following limit : limx→0[5x+33-2x]2x - Mathematics and Statistics

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प्रश्न

Evaluate the following limit : 

`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/x)`

योग

उत्तर

`lim_(x -> 0)[(5x + 3)/(3 - 2x)]^(2/x)`

= `lim_(x -> 0) [(3 + 5x)/(3 - 2x)]^(2/x)`

= `lim_(x -> 0) [(1 + (5x)/3)/(1 - (2x)/3)]^(2/x)`

= `lim_(x -> 0)((1 + (5x)/3)^(2/x))/((1 - (2x)/3)^(2/x))`

= `(lim_(x -> 0) (1 + (5x)/3)^(2/x))/(lim_(x -> 0) (1 - (2x)/3)^(2/x))`

= `(lim_(x -> 0)[(1 + (5x)/3)^(3/(5x))]^(10/3))/(lim_(x -> 0)[(1 - (2x)/3)^((-3)/(2x))]^((-4)/3)`

= `([lim_(x -> 0) (1 + (5x)/3)^(3/(5x))]^(10/3))/([lim_(x -> 0) (1 - (2x)/3)^((-3)/(2x))]^((-4)/3)`

= `("e"^(10/3))/("e"^((-4)/3))   ...[(because x -> 0  therefore (5x)/3 -> 0"," (-2x)/3 -> 0),(and lim_(x -> 0) (1 + x)^(1/x) = "e")]`

= `"e"^(14/3)`.

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Limits of Exponential and Logarithmic Functions
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Exercise 7.6 [पृष्ठ १५४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 7 Limits
Exercise 7.6 | Q II. (3) | पृष्ठ १५४

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