IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

naman

_{[x ]}

_{[o ]} [sin²(

/2-px)]^{sec^2(}

/2-qx)

iitkgp_bipin

sin(

/2 - px) = cospx
sec(

/2 - qx) = cosecqx

L = _[x]

_[0] {cos²px}^{(cosecqx)^2}

As x tends to 0 , cospx tends to 1 and cosecqx tends to infinity.
Hence the limit is in the form of 1^infinity.
So apply , L = _[x]

_[0] {f(x)}^g(x) = _[x]

_[0] exp[{f(x)-1}g(x)]

Here {f(x)-1}g(x) = (cos²px - 1)(cosec²qx) = (-sin²px) / (sin²qx)

Now _[x]

_[0] (-sin²px) / (sin²qx) = -p²/q²

Hence L = e^{-(p^2/q^2)}

punterjack

Lt can be simpified to

(1-sin2px)^1/sin2qx=(1-p2x2)^1/q2x2as as lt x->0 sin ax->ax

multiplying and dividing the power by -1,

and multiplying and dividing the by (q2/p2) we have Ltf(x)=

e^-(p2/q2)

Please rate me if u found my answer satisfactory

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