If a, b and c are real numbers, then the value of \(\displaystyle\lim_{t\to 0}\ln\left(\frac{1}{t}\int_{0}^{t}(1+a\sin(bx))^{c/x}\,dx\right)\) equals
abc
Let \(g(x)=(1+a\sin(bx))^{c/x}\). As \(x\to 0\), \(\sin(bx)\sim bx\), so \(g(x)=\left[(1+abx+o(x))\right]^{c/x}\to e^{abc}\) (standard limit \((1+kx)^{1/x}\to e^{k}\) with \(k=ab\), raised further to the power \(c\)).
The inner quantity \(\dfrac1t\displaystyle\int_0^t g(x)\,dx\) is of the form \(0/0\) as \(t\to 0\), so L'Hopital's rule applies: differentiating numerator and denominator with respect to \(t\) (using the Fundamental Theorem of Calculus on the numerator) gives \(\displaystyle\lim_{t\to0}\dfrac1t\int_0^t g(x)\,dx=\lim_{t\to0}g(t)=e^{abc}\).
So the average value inside the logarithm tends to \(e^{abc}\), a fixed positive number.
Since \(\ln\) is continuous, the whole limit equals \(\ln\!\left(e^{abc}\right)=abc\).
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