JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 3)
$$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$ is equal to
$$ - {1 \over 5}$$
$$ - {1 \over 10}$$
0
$$ {1 \over 10}$$
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$
This is in $${0 \over 0}$$ form.
So apply newton leibniz rule
$$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$$ = 0
This is in $${0 \over 0}$$ form.
So apply newton leibniz rule
$$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$$ = 0
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