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प्रश्न
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
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उत्तर
Let f(x) = `lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) |
`(sqrt(3 - 0.1) - sqrt(3))/(- 0.1)` = `(1.703 - 1.732)/(- 0.1)` = 0.29 |
`(sqrt(3 - 00.1) - sqrt(3))/(- 0.01)` = `(1.703 - 1.732)/(- 0.01)` = 0.288 |
`(sqrt(3 - 0.001) - sqrt(3))/(- 0.001)` = `(0.000284)/(- 0.001)` = 0.288 |
`(sqrt(3 + 0.001) - sqrt(3))/(- 0.001)` = `(0.000289)/(0.001)` = 0.289 |
`(sqrt(3 + 0.01) - sqrt(3))/(0.01)` = `(0.00288)/(0.01)` = 0.288 |
`(sqrt(3 + 0.1) - sqrt(3))/(0.1)` = `(0.0286)/(0.1)` = 0.286 |
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x` = 0.288
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