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In problems 1 – 6, using the table estimate the value of the limitlimx→2x-2x2-4

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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
तक्ता
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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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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९५]

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

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