lim x->0 (sinx/x)^(1/x^2) Solution follow this L = lim (x-->0+) [sin(x)/x]^(1/x^

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follow this L = lim (x-->0+) [sin(x)/x]^(1/x^2) ==> ln(L) = lim (x-->0+) ln[sin(x)/x]/x^2 = lim (x-->0+) {ln[sin(x)] - ln(x)}/x^2. Applying L'Hopital's Rule yields: lim (x-->0+) {ln[sin(x)] - ln(x)}/x^2 = lim (x-->0+) [cos(x)/sin(x) - 1/x]/(2x) = lim (x-->0+) [x*cos(x) - sin(x)]/[2x^2*sin(x)]. Applying L'Hopital's Rule again gives: lim (x-->0+) [x*cos(x) - sin(x)]/[2x^2*sin(x)] = lim (x-->0+) [cos(x) - x*sin(x) - cos(x)]/[4x*sin(x) + 2x^2*cos(x)], by the product rule = lim (x-->0+) [-x*sin(x)]/[4x*sin(x) + 2x^2*cos(x)]. Dividing the numerator and denominator by x^2: lim (x-->0+) [-x*sin(x)]/[4x*sin(x) + 2x^2*cos(x)] = lim (x-->0+) [-sin(x)/x]/[4sin(x)/x + 2cos(x)] = -1/(4 + 2), since lim (x-->0) sin(x)/x = 1 = -1/6. Thus, ln(L) = -1/6 and so L = lim (x-->0+) [sin(x)/x]^(1/x^2) = e^(-1/6).

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