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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

In problems 1 – 6, using the table estimate the value of the limitlimx→0x+3-3x 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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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
सारिणी
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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९५]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 3 | पृष्ठ ९५

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