Let $f:R o \left( 0,\infty ight)$ be such that $f\left( x ight)+\frac{e^{x+x^{2}}}{f\left( x

Question

Let $f:R	o \left( 0,\infty ight)$ be such that $f\left( x ight)+\frac{e^{x+x^{2}}}{f\left( x ight)}\le e^{x}+e^{x^{2}}, \forall$ $x>0$. Then, $\displaystyle\lim_{x 	o 1} f\left( x ight)$ is

Answer

$(f(x))^{2}-\left(e^{x}+e^{x^{2}}ight)f(x)+e^{x}.e^{x^{2}}\le 0$ $\implies \left(f(x)-e^{x}ight)\left(f(x)-e^{x^{2}}ight)\le 0$ $\implies e^{x^{2}}\le f(x)\le e^{x}\ \forall\ x\in(0,1)$ $\implies e^{x}\le f(x)\le e^{x^{2}}\ \forall\ x\in(1,\infty)$ By Sandwich Theorem, $\displaystyle\lim_{x	o 1}(e^{x})=\lim_{x	o 1}(e^{x^{2}})=e$

Practice More