IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - find the limit

spideyunlimited

_{[ x ]}

_[ 0 ] ( sin x / x ) ^{(sin x / x - sinx )} = ?

i think theres something to do like bringing it to 1 ^inf form... plz highlight that too

krishna.gopal

sin(x) = x-x³/3! +x⁵/5! plus higher order terms

sin(x)/x= 1-x²/ 6+x⁴/120 plus higher order terms

sin(x)/(x-sin(x))= 1/(1-sin(x)/x)= 1/(x²/ 6-x⁴/120 plus higher order terms) = 6/x²

So given limits becomes

_{[x ]}

_{[0 ]} (1-x²/6)^6/x2=e^-1

joyfrancis

The answer is inf.

Solution:

sinx/x when x tends to 0 = 1 {tending to}

sinx/x-sinx tends to infinity when x tends to 0

apply lh

=>cosx/1-cosx

put x=0

limit =inf.

.: we have 1^inf form

let the expression be h(x)

h(x)=e^(sinx/x)(2sinx-x/x-sinx)

=e^(2sin^2x-xsinx/x^2-xsinx)

We just have to find

limx-->0 2sin^2x-xsinx/x^2-xsinx

it is easy now,

since it is 0/0 form apply lh

it becomes =>(2sin2x-xcosx-sinx)/(2x-xcosx-sinx)

again this is 0/0 so again differentiate

you willl get

(4cos2x+xsinx-2cosx)/(2+xsinx-2cosx)

Put x=0

4+0-2/2+0-2

=2/0

=inf

We thus have the answer as e^inf=inf

priyesh

the ans is e^-1

_{x ][ 0 ]} ( sin x / x ) ^{(sin x / x - sinx )}

first we convert the limit to 1^inf form so

= _{x ][ 0 ]} ( sin x / x ) ^{(1/ (x/sinx - 1))} (dividing the num & denom. of the power by sinx)

= (1 + (sinx/x - 1))^{(1/(sinx/x -1) * (sinx/x -1)/(x/sinx - 1)}

now the limit of the bold part becomes e because (lim x---->0 (1 + x)^1/x = e)

now the lim becomes e^{(lim x--->0 (sinx/x -1)/(x/sinx - 1)}

= e^{(lim x--->0 (-sinx/x))}

= e^-1

hope you understood

cheers

khinart

priyesh is absolutely correct.The answer is e^-1

spideyunlimited

thanks @krishnagopal . your method was quite innovative u used the sin series. thanks :)

well priyesh got wht i was talkin about though. gud job priyesh .. i had done the question halfway but wasnt able to proceed.. theres one more with tan in it and u get e square as the answer tht as easy.

spideyunlimited

@joyfrancis.. man wht r u blundering.. lol... we dont get infinity.

Ask & Discuss Questions with Community & Experts