IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

2008sIITian

Evaluate

_{[ x]}

_{[ a]} a^xsin(b/a^x) when 0<a<1&when a>1

naveen_nvn

since it is not in any indeterminant form....
why not the answer just b...a^a sin(b/(a^a)) in both the cases...???

iitjee08aspirant

i think in the case when a>1 the expression b/a^x will become small close to zero so acc to me the ans for 2nd case should be 0
plzzz tell me if i am wrong

iitkgp_bipin

_[x]

_[a] a^xsin(b/a^x) when 0<a<1 & when a>1

Limit can be evaluated by directly putting x=a since the limit is not in indeterminate form.

Hence the limit comes out to be a^asin(b/a^a)

partha

when 0<a<1 the ans is sin(b)

when a>1 the ans is a^[a] sin (b/a^[a])
please explain if i am wrong

partha

is x tending to greatest integer of a???

him26.89

look expansion of siny is

siny = y - y^3/3! +y^5/5! .................

when a>1 then ans is

b - b^3/3!...... = sinb

& for 0<a<1, ans is (-)infinity.

bcos a^xwill tend to infinity. (u can evaluate it separately for urself)

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