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Question

Find the limx0(x+4)3/2+ex9x without using L'Hospital rule

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Solution

(x+4)3/2=8(1+x4)3/2

(1+x/m)n=1+nxm+n(n1)2!x2m2+.....

Hence expansion of (1+x4)3/2 =1+32x +(32)(12)2!x2.....

now expansion of ex=1+x1!+x22!+.....

now put these expansion on given expression

limx0(x+4)3/2+ex9x =8(1+32x4+38x242)+(1+x1!+x22!)9x

On solving this equation we will get
limx0(x+4)3/2+ex9x=4

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