# Supremum of $\underset{n \to \infty }{\lim} \underset{x\in [0,1]}{\sup} \left | \frac{x+x^{2}}{1+n+x} \right |$

I need to find $$\underset{n \to \infty}{\lim} \underset{x\in [0,1]}{\sup} \left| \frac{x+x^{2}}{1+n+x} \right|.$$ How to show that supremum will be at the point $x=1$?

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Then you could take a derivative, sett it equal to zero to find your critical points, etc. This answers your question of deciding for which $x$ it attains its max.
On the other hand, you could be generous and say the numerator is at most $2$, and the denominator is at least $1+n$. This is easier and sufficient to solve the supremum problem.
First let us find the supremum of $f_n(x) = \dfrac{x+x^2}{1+n+x}$. We have $$f_n'(x) = \dfrac{(1+n+x)(1+2x) - (x+x^2)}{(1+n+x)^2} = \dfrac{x(2n+x+2)+n+1}{(1+n+x)^2} > 0 \,\,\,\, \forall x \in [0,1]$$ Hence, $f_n(x)$ is an increasing function in the interval $[0,1]$. Hence, the supremum is attained at $x=1$. We hence get that $$\sup_{x \in [0,1]} f_n(x) = \dfrac2{n+2} \implies \lim_{n \to \infty} \sup_{x \in [0,1]} f_n(x) = \lim_{n \to \infty} \dfrac2{n+2} = 0$$