LIMITS:

1) as f is odd function f(-x)=-f(x) or f(-h)=-f(h) ..........e.q.1

now as limit exists --- l.h.l = r.h.l.

lt_x--0^- f(x) = lt_x--0⁺ f(x)

a) let x=o- h

then, l.h.l. = lt_h--0 f(-h) .....e.q.2

b) let x 0+h

then r.h.l. = lt_h--0 f(h).........e.q.3

now l.h.l. = r.h.l.

lt_h--0 f(-h) = lt_h--0 f(h).........from eq.2 and eq.3

so, f(-h)= f(h).........and,

- f(-h)=f(h) .....................from e.q.1

adding these

f(h)=0 hence,

lt_h--0 0 = 0

therefore , option A should be correct......

3) lt _n---∞ = 5ⁿ⁺¹+3ⁿ-2²ⁿ5ⁿ+2ⁿ+3²ⁿ⁺³

numerator = 5ⁿ.3ⁿ.2²ⁿ{53ⁿ.2²ⁿ + 15ⁿ.2²ⁿ-1 5ⁿ.3ⁿ}

denominator = 5ⁿ.3²ⁿ.2²ⁿ{13²ⁿ.2²ⁿ + 2^-n5ⁿ.3²ⁿ+27 5ⁿ.2²ⁿ}

lt _n---∞ =numeratordenominator=5ⁿ.3ⁿ.2²ⁿ{53ⁿ.2²ⁿ + 15ⁿ.2²ⁿ-1 5ⁿ.3ⁿ}
5ⁿ.3²ⁿ.2²ⁿ{13²ⁿ.2²ⁿ + 2^-n5ⁿ.3²ⁿ+27 5ⁿ.2²ⁿ}

lt _n---∞ ={53ⁿ.2²ⁿ + 15ⁿ.2²ⁿ-1 5ⁿ.3ⁿ}.3ⁿ{13²ⁿ.2²ⁿ + 2^-n5ⁿ.3²ⁿ+27 5ⁿ.2²ⁿ}.3ⁿ.3ⁿ

=1∞=0

hence option D is correct......

Why are u assuming that limit exists?Can't it be none of these?

4) lt_x--0 x³√a+x.(bx-sinx)

applying l hospital rule...

lt_x--0 3x².2√a+xbx-sinx+2(a+x)[b-cosx]

=02a[b-1]=1 (given)

it should be converted in 0/0 form....

a cannot be equal to 0 as a>0 given.so, [b-1]=0

hence b=1

now,

again applying l hospital rule....

lt_x--0 24x√a+x+6x²[2√a+x].2+2asinx-2cosx+2xsinx+1-cosx =0/0

again applying l hospital rule...

lt_x--0 24√a+x++24x2√a+x+12x[2+2asinx-2cosx+2xsinx+1-cosx]√a+x+ 2√a+x{2acosx+2sinx+2xcosx+3sinx}

=24√a2√a{2a+0+0+0+0}

122a=1

a=6

hence b=1 and a=6.........so,d is correct

Thanx for the solutions. Although u have given way too much xplanations, but its ok. Although the answer to ques no 4 is given as (A), but i will check with dat. Anyway thanx again. I've posted 2 more doubts in limits. i'll b happy if u can solve them too for me. But 2 save ur time and effort u can cut down on those big xplanations(especially da one dat u gave in ques no 1). But minilmal xplanation is reqd as i m nt dat a hi-fi student!!!...
Bye. Hey btw are u an IItian??