LIMITS

2.) (n+1)¹⁰+(n+2)¹⁰+..........+(n+100)¹⁰
limn->∞ --------------------------------------------------------------------------------
n¹⁰ +10¹⁰

(10C0n¹⁰+10C2n⁹...)+(10C0n¹⁰2⁰ +...) +... 100 terms
= limn->∞ --------------------------------------------------------------------------------
n¹⁰ +10¹⁰

100X10C⁰ + 0
= limn->∞ --------------------------------------------------------------------------------
1+0

= 100

Solve 1... please.. i could not understand how.. :(

1st one, for aneesh... :

n²+1 / n+1 - an

= n²-1+2 / n+1 - an

= n-1 + 2/n+1 - an

lim(n→∞) [(1-a)n - 1 + (2/n)/1+(1/n)]

lim(n→∞) [(1-a)n] - 1 + 0

for d above limit to be finite, coefficient of n must be =0 => a=1

[1]

first is..
n²(1-a²) -an +1-a
-------------------------------
(n+1)

dividing numerator and dinominator we have n left only in numerator as n(1-a²)

so for limit to exist (1-a²)=0
or a=±1