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For all $x∈ (0,1)$ |
$e^x<1+x$ $\log_e(1+x)<x$ $\sin x>x$ $\log_e x>x$ |
$\log_e(1+x)<x$ |
$e^x=1+x+\frac{x^2}{2!}+.....>1+x⇒x>\log_e(1+x)$ i.e., $\log_e(1+x)<x$ |
