Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x=1$?

share|improve this question

2 Answers 2

It's positive, so you don't need absolute values.

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.

share|improve this answer

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 $$\sup_{x \in [0,1]} f_n(x) = \dfrac2{n+2} \implies \lim_{n \to \infty} \sup_{x \in [0,1]} f_n(x) = \lim_{n \to \infty} \dfrac2{n+2} = 0$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.