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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 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
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उत्तर
`lim_(x -> 2) (x - 2)/(x^2 - 4) = lim_(x -> 2) (x - 2)/((x + 2)(x - 2))`
= `lim_(x -> 2) 1/((x + 2))`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) |
`1/(1.9 + 2)` = `1/3.9` = 0.256 |
`1/(1.99 + 2)` = `1/3.99` = 0.251 |
`1/(1.999 + 2)` = `1/3.999` = 0.250 |
`1/(2.001 + 2)` = `1/4.001` = 0.249 |
`1/(2.01 + 2)` = `1/4.01` = 0.249 |
`1/(2.1 + 2)` = `1/4.1` = 0.244 |
`lim_(x -> 2) (x - 2)/(x^2 - 4) = lim_(x -> 2) 11/(x + 2)`
= `1/(2 + 2)`
= `1/4`
= 0.25
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