1. Evaluate limln{(x2 + 1)1/2- x} / x (where x tends 0)
2. Evaluate lim n2{(a)1/n - (a)1/1+n} (where n tends to infinity) ( a >0 , n=N)

Hi Sashant,
Please find below the solution to the asked query:

L=limx0  ln1+x2-xxOn putting x=0, both numerator and denominator become 0.It is 00 form. Hence L'Hopital rule can be applied,Hence differentiating numerator and denominator separately we get,L=limx0  11+x2-x121+x2.2x-11=limx011+x2-xx1+x2-1=limx011+x2-x1+x2+x1+x2+xx1+x2-1=limx01+x2+x1+x2-x2x1+x2-1=limx01+x2+x1x1+x2-1Applying limit we get,L=1+0+0101+0-1L=-1limx0  ln1+x2-xx=-1 Answer

L=limn n2a1n-a11+n=limn a11+n.n2a1n-11+n-1Now limn a11+n=a11+=a0=1L=limn n2a1n-11+n-1=limn n2a1nn+1-1=limn n2a1nn+1-1nn+1nn+1=limn n2nn+1a1nn+1-11nn+1=limn n2n2+na1nn+1-11nn+1Note: In  form,Answer=Coefficient of highest degree term of the ploynomial in numeratorCoefficient of highest degree term of polynomial in denominatorlimn n2n2+n=Coefficient of n2 in numeratorCoefficient of n2 in denominator=11=1L=limn a1nn+1-11nn+1Here limn1nn+1=1=0By standard form we know thatlimxa afx-1fx=lna where limxafx=0 Note: Make sure that fx in the power and in denominator are exactly samelimn a1nn+1-11nn+1=lnaHere L=limn n2a1n-a11+n=a Answer

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