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If lim(xto0)(f(x))/(sin^(2)x)=8,lim(xto0...

If `lim_(xto0)(f(x))/(sin^(2)x)=8,lim_(xto0)(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda" and "`
`lim_(x to 0)(1+2f(x))^((1)/(g(x)))=(1)/(e)," then"`
`lim_(x to 0)(1+f(x))^((1)/(2g(x)))` is equal to

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The correct Answer is:
A

`underset(xto0)lim(f(x))/(sin^(2)x)=8`
`implies" "underset(xto0)lim((f(x))/(x^(2)))/((sin^(2)x)/(x^(2)))=8`
`implies" "underset(xto0)lim(f(x))/(x^(2))=8" "...(1)`
Also, `underset(xto0)lim(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda`
`implies" "underset(xto0)lim((g(x))/(x^(2)))/((-2(1-cosx))/(x^(2))-(e^(x)-1)/(x)+x)=lamda`
`implies" "underset(xto0)lim((g(x))/(x^(2)))/((-4"sin"^(2)(x)/(2))/(x^(2))=-1+0)=lamda`
`implies" "underset(xto0)lim(g(x))/(x^(2))=-2lamda" "...(2)`
Now, `underset(xto0)lim(1+2f(x))^((1)/(g(x)))`(one power infinity form)
`=e^(underset(xto0)lim(2f(x))/(g(x)))=e^(underset(xto0)lim(2(f(x)//x^(2)))/((g(x)//x^(2))))`
`=e^(underset(xto0)lim(16)/(-2lamda))`
`=e^(-(8)/(lamda))`
`e^(-1)`(given)
`:." "lamda=8`
`underset(xto0)lim(1+f(x))^((1)/(2g(x)))=e^(underset(xto0)lim(f(x))/(2g(x)))`
`=e^(underset(xto0)lim((f(x)//x^(2)))/(2(g(x)//x^(2))))`
`=e^(underset(xto0)lim(8)/(2xx(-16)))`
`=e^(-(1)/(4))`
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