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प्रश्न
In problems 1 – 6, using the table estimate the value of the limit.
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.344820 | 0.33444 | 0.33344 | 0.333222 | 0.33222 | 0.332258 |
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उत्तर
`lim_(x -> 2) (x - 2)/(x^2 - x - 2) = lim_(x -> 2) ( x - 2)/((x - 2)(x + 1))`
= `lim_(x -> 2) (x - 2)/(x + 1)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 20.1 | 2.1 |
| f(x) |
`1/(1.9 + 1)` = `1/2.9` = 0.34 |
`1/(1.99 + 1)` = `1/2.99` = 0.33 |
`1/(1.999 + 1)` = `1/2.99` = 0.33 |
`1/(2.001 + 1)` = `1/3.001` = 0.33 |
`1/(2.01 + 1)` = `1/3.01` = 0.33 |
`1/(2.1 + 1)` = `1/3.1` = 0.32 |
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)` = 0.3
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