DOUBT 3

BITSAT 2015 Forum › Calculus questions › DOUBT 3

The value of lim_x-->0(cosx + a sinbx)^1/x

Ans was e^ab

The method was using the expansion...but pls explain how the higher powers gets cancelled out..(because denominator of cos and sin goes as 2! , 3! respectively)

#Mathematics #Calculus

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4 Answers

msp ·2009-05-11 08:49:55

if u want to find the soln for ur dbt substitute for many nos.and u will get a conclusion.

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sriraghav ·2009-05-11 08:51:21

but it is not the proper method na....U can do in some cases but not always.....(especially calculus might mislead u)

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Celestine preetham ·2009-05-11 08:56:22

ans is

put cosx =1
sinbx= bx

(1+abx)^1/x = e^log(1+abx)/x = e^ab

ive skipped the word limit

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Mani Pal Singh ·2009-05-11 09:02:50

u simply have to observe that it is 1^âˆž form
so

use the standard form

e^1/x^{(cosx + a sinbx)}

so now here use expansions
@ cele we should not promote the putting thing when we r at home [1]

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DOUBT 3

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