A) 0
B) 1
C) \[\frac{1}{2}\]
D) non-existent
Correct Answer: C
Solution :
\[\underset{x\to \infty }{\mathop{Lim}}\,\frac{1}{{{a}^{2}}}\,\int\limits_{0}^{a}{\ln \,(1+{{e}^{x}})\,dx,}\] we get Using Leibnitz Rule \[\underset{a\to \infty }{\mathop{Lim}}\,\,\frac{\ln (1+{{e}^{a}})}{2a}\,\left( \frac{\infty }{\infty } \right)=\frac{1}{2}\,\underset{x\to \infty }{\mathop{Lim}}\,\,\frac{{{e}^{a}}}{1+{{e}^{a}}}\,=\frac{1}{2}\]You need to login to perform this action.
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