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

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

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

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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