IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

layman

1.Evaluate

$lim_{x o 0}left( left[ rac{nsin(x )}{x} ight] + left[ rac{n an(x )}{x} ight] ight).$

where [ ] is the greatest integer function and n is an integer

layman

Of course without L-Hospital's Rule.

computer001

1) n-1 + n
= 2n-1
sinx/x sltely less than 1 tanx/x slightly gr8er than 1 as x->0

layman

Yes right.

sboosy

computer001

1/3
write as lim t->0 ({(1+t+t^3)^1/3 -1}/1+t+t^3-1}*1+t^2
gives 1/3*(1)^-2/3 *1

hash_include

2.
$lim$ $x(1 + rac 1 x + rac 1 {x^2})^{ rac 1 3} - x$

$= lim$ $rac {( rac 1 x + rac 1 {x^2})^{ rac 1 3} - 1}{ rac 1 x}$

apply Lhospital

$rac { rac 1 3( rac 1 x + rac 1 {x^2})^{ rac {-2} 3})( rac {-1}{x^2} + rac {-3}{x^4})}{ rac {-1} {x^2}}$

solving we get answer = $rac 1 3$
Smile

EDIT: sorry dint see that 'without lhospital'

computer001

i think q says widout L'hospital :\

layman

No LH hash-include

layman

computer001 can u pls explain it more clearly.

computer001

2)
put 1/x = t
then we get lim t->0 {1+1/t^2 +1/t^3}^1/3 -1/t
= {{1+t+t^3}^1/3 -1}/t

multiply and devide by t^2 + 1
so v have
{{1+t+t^3}^1/3 -1}(t^2+1)/{t(t^2+1)}

now denom write as t^3+t +1-1
put t^3+t+1 as y so y->1

so v have lim y->1 (y)1/3 -1/y-1 * lim t->0 (1+t^2)

so 1/3*1=1/3

sboosy

computer001

sboosy..no L'hospital

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