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Question

The value of limxπ2-tan-1x1x is


A

0

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B

1

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C

-1

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D

e

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Solution

The correct option is B

1


Explanation of the correct option.

Step 1: Put the values of limit.

Given : limxπ2-tan-1x1x

Let L=limxπ2-tan-1x1x,

taking ln both sides,

lnL=limxlnπ2-tan-1x1x

lnL=limx1xlnπ2-tan-1xlnL=limx1xlncot-1xlnL=

Step 2: Apply L.Hospital's rule.

Since, it is form, applying L.Hosptal's rule,

lnL=limx1cot-1x-11+x21

lnL=-limx1+x2-1cot-1xlnL=00

Since, it is 00 form, applying L.Hosptal's rule,

lnL=limx-2x(1+x2)2-11+x2

lnL=-2limxx1+x2lnL=

Since, it is form, apply L.Hosptal's rule,

lnL=-2limx12x

lnL=-2(0)lnL=0L=e0L=1

Therefore the value of limxπ2-tan-1x1x is 1.

Hence, option (B) is the correct option, i.e. 1.


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