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Evaluate the following limits: eelimx→0ex-e-xsinx - Mathematics

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प्रश्न

Evaluate the following limits:

`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`

योग
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उत्तर

We know `lim_(x -> 0) ("e"^x - 1)/x` = 1

`lim_(x -> 0) sinx/x` = 1

`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx =  lim_(x -> 0) ("e"^x - 1/"e"^x)/sinx`

= `lim_(x -> 0) (("e"^x * "e"^x - 1)/"e"^x)/(sinx)`

= `lim_(x -> 0) ("e"^(2x) - 1)/("e"^x sinx)`

= `lim_(x -> 0) (1/"e"^x xx ("e"^(2x) - 1)/(1/2 xx 2x) xx x/sinx)`

= `(lim_(x -> 0) 1/"e"^x) 2(lim_(2x -> 0) ("e"^(2x) - 1)/(2x)) xx 1/((lim_(x -> 0) sinx/x))`

= `1/"e"^0 xx 2 xx 1 xx 1/1`

= `1/1 xx 2 xx 1`

`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx` = 2

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Concept of Limits
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 25 | पृष्ठ ११८

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