Mathematics

If $$a,b$$ and $$c$$ are real numbers then the value of $$\mathop {\lim }\limits_{t \to 0} {l_n}\left( {\frac{1}{t}\int_0^1 {{{\left( {1 + a\sin bx} \right)}^{\frac{c}{x}}}dx} } \right)$$ equals


ANSWER

$$abc$$


SOLUTION
Let $$L={\lim}_{t\rightarrow\,0}\ln{\left(\dfrac{1}{t}\displaystyle\int_{0}^{t}{{\left(1+a\sin{bx}\right)}^{\frac{c}{x}}dx}\right)}$$
It is of $$\dfrac{0}{0}$$ form.
Hence we use L'Hospitals's rule
$$L={\lim}_{t\rightarrow\,0}\dfrac{{\left(1+a\sin{bt}\right)}^{\frac{c}{t}}\times 1}{1}$$
$$L={\lim}_{t\rightarrow\,0}\ln{{e}^{a\sin{bt}\times\dfrac{c}{t}}}$$
$$L={\lim}_{t\rightarrow\,0}\dfrac{ac\sin{bt}}{t}$$ is of $$\dfrac{0}{0}$$ form.
$$L={\lim}_{t\rightarrow\,0}\dfrac{ac\cos{bt}}{1}$$
$$=abc$$
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Single Correct Hard Published on 17th 09, 2020
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