मराठी

lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]

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प्रश्न

`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`

बेरीज
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उत्तर

`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`

Step 1: Multiply and divide by the conjugate

`x(sqrt(x^2+1)-sqrt(x^2-1)) xx (sqrt(x^2+1)+sqrt(x^2-1))/(sqrt(x^2+1)+sqrt(x^2-1))`

`x xx ((x^2+1) - (x^2-1))/(sqrt(x^2+1) + sqrt(x^2-1))`

(x2 + 1) − (x2 − 1) = 2

`= (2x)/(sqrt(x^2+1)+sqrt(x^2-1)`

Step 2: Factor out xxx from the denominator

`= (2x)/(xsqrt(1+1/x^2) + xsqrt(1-1/x^2))`

`= (2x)/(xsqrt(1+1/x^2) + sqrt(1-1/x^2))`

`= (2)/(sqrt(1+1/x^2) + sqrt(1-1/x^2))`

Step 3: Take the limit as x → ∞

`1/x^2 -> 0`

`sqrt(1+1/x^2) ->1, sqrt(1-1/x^2) ->1`

`2/(1+1)`

`2/2`

= 1

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पाठ 29: Limits - Exercise 29.6 [पृष्ठ ३९]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.6 | Q 12 | पृष्ठ ३९

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