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PIYUSH AGRAWAL Grade: 12
        

Q.1.limx-->infinity(x+2)tan-1(x+2)-xtan-1x


Q.2.limx-->0[{(tan-1x)/x}-{(sin-1x)/x}]1/2


Q.3.limx-->0[{(sin-1x)/x}-{(tan-1x)/x}]1/2

6 years ago

Answers : (1)

Vijay Luxmi Askiitiansexpert
357 Points
										

Dear Piyush,


For Q1  split (x+2) term. and use suitable inverse trig sum of tan inverse alse as x--->infinity  that means x+2 also---> infinity. and get the ansver.

For both Q2 and Q3 we get 0/0 form so use L'Hospital rule.


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Vijay Luxmi


6 years ago
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