Q 1 ) lim x----∞ (√ x √ x + √x - √ x ) Q 2) lim x----∞ x 3/2 √ x 3 + 1 - √ x 3 - 1 ans : 1 Q 3) lim x----∞ 1/ 1-n 4 + 8 / 1-n 4 + ........... + n 3 / 1-n 4 ans: - 1 / 4 ?

Question

Q 1 ) lim x---->&infin; (&radic; x &radic; x + &radic;x - &radic; x ) Q 2) lim x---->&infin; x 3/2 &radic; x 3 + 1 - &radic; x 3 - 1 ans : 1 Q 3) lim x---->&infin; 1/ 1-n 4 + 8 / 1-n 4 + ........... + n 3 / 1-n 4 ans: - 1 / 4 ?

Jack sparrow · Answer

the answer is infinity

Swapnil Saxena · Answer

Ans 3. (1+ 8 + 27 + ... + n 3 )/1-n 4 Since 1 3 +2 3 +3 3 +...+n 3 = (n(n+1)) 2 /4 then the expression becomes (n(n+1)) 2 /4(1-n 4 ) = ((n) 2 (n+1) 2) /4(1-n 4 ) dividing the whole expression ny n 4 . Now the expression is =(1)(1+1/n 2 ))/4(1/n 4 -1) = No since n->infinity 1/n tends to 0 thus byu using direct substitutions = 1/-4 =-1/4

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Q 1 ) lim x---->∞ (√ x √ x + √x - √ x ) Q 2) lim x---->∞ x 3/2 √ x 3 + 1 - √ x 3 - 1 ans : 1 Q 3) lim x---->∞ 1/ 1-n 4 + 8 / 1-n 4 + ........... + n 3 / 1-n 4 ans: - 1 / 4 ?

Q 1 ) lim_x---->∞(√ x √ x + √x - √ x )

Q 2) lim_x---->∞x^3/2 √ x³ + 1 - √ x³ - 1

ans : 1

Q 3) lim_x---->∞ 1/ 1-n⁴ + 8 / 1-n⁴ + ........... + n³ / 1-n⁴

ans: - 1 / 4

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