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In problems 1 – 6, using the table estimate the value of the limitlimx→-31-x-2x+3 x – 3.1 – 3.01 – 3.00 – 2.999 – 2.99 – 2.9 f(x) – 0.24845 – 0.24984 – 0.24998 – 0.25001 – 0.25015 – 0.25158

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प्रश्न

In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`

x – 3.1  – 3.01 – 3.00 – 2.999 – 2.99 – 2.9
f(x) – 0.24845 – 0.24984 – 0.24998 – 0.25001 – 0.25015 – 0.25158
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उत्तर

Let f(x) = `lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`

x – 3.1 – 3.01 – 3.00 – 2.999 – 2.99 – 2.9
f(x)

`(sqrt(1 + 3.1) - 2)/(- 3.1 + 3)`

= `(0.0248)/( - 0.1)`

= – 0.248

`(sqrt(1 + 3.01) - 2)/(- 3.01 + 3)`

= `(0.00250)/( - 0.01)`

= – 0.249

`(sqrt(1 + 3) - 2)/(- 3 + 3)`

= `0/0`

= 0

`(sqrt(1 + 2.999) - 2)/(- 2.999 + 3)`

= `(- 0.000250)/(0.001)`

= – 0.25

`(sqrt(1 + 2.99) - 2)/(- 2.99 + 3)`

= `(- 0.00250)/(0.01)`

= – 0.25

`(sqrt(1 + 2.9) - 2)/(- 2.9 + 3)`

= `(- 0.02515)/(0.1)`

= – 0.2515

`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)` = – 0.25

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९५]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 4 | पृष्ठ ९५

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