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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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