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We are given $g(x)=\frac{x \sin x}{x+1}$, and as I said we need to show it has no maxima in $(0,\infty)$.

My attempt: assume there is some $x_0>0$ that yields a maxima. then for all $x$

$$-1+\frac{1}{x+1}\leq \frac{x \sin x}{x+1}\leq \frac{x_0 \sin x_0}{x_0+1}\leq 1-\frac{1}{x_0+1}$$ and we can find some $x$ for which this isn't satisfied (like $\frac{1}{2-\frac{1}{x_0+1}}$?). This feels very unnecessary (plus I assume things about $x_0$), but I don't know what good way there is...

Note that I saw the thread here showing $\sup_{x>0} g(x)=1$ but one solution is not on my level, and the other one doesn't actually show its the $\sup$, rather than some sort of "partial limit". Even so, I still don't know how to show that there is no $x$ such that $g(x)=1$, and I don't think it's needed here.

Any help is appreciated in advance!

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    Can you show that $g(x) < 1$ for all $x$? Can you show that $g(x)$ comes arbitrarily close to $1$, however? – 2017-01-09

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Notice that, for $x_n=\frac{(4n+1)\pi}{2}n$, $n\in\mathbb N$, $\sin(x_n)=1$, so $f(x_n)=\frac{x_n}{x_n+1}$.

We also have $f(x)<1$ for all $x\in(0,+\infty)$. The sequence $f(x_n)$ is strictly increasing and gets closer and closer to $1$, so that means $f$ cannot have a maxima.

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    Welcome to the website first of all! And how come you drew the last conclusion, that $f$ cannot have a maxima? at $x_n$ you're not looking at all points. Between two points the function can "behave" in any way it wants, so we can have a maxima there. (I hope you catch my drift) – 2017-01-09
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    Thanks! All right, it could behave in any way (actually it doesn't), but if it were a point of those you mention, say, $\alpha$, it has to be $f(\alpha)<1$, so we can go further in the sequence $f(x_n)$ and find some $x_N$ that $f(x_N)>f(\alpha)$, as $f(x_n)$ gets as close to $1$ as we want. – 2017-01-09
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    Actually, if there were a number $\alpha$ that satisfies $f(\alpha)> f(x_n)=\frac{x_n}{x_n+1}$ for all $n$, it has to be $f(\alpha) \geq \lim_ {n\to\infty} \frac{x_n}{x_n+1}=1$, which cannot be, since $f(x)<1$ for all $x$. – 2017-01-09
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    Can we use the limit definition for $\epsilon=1-f(\alpha)$ to get that theres a $N$ that for $n>N$, $|f(x_n)-1|=1-f(x_n)<1-f(\alpha)$? – 2017-01-09
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    Yes, we can use that, of course~ – 2017-01-09
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for the first derivative we get $$\frac{(x^2+x)\cos(x)+\sin(x)}{(x+1)^2}$$ but this derivative hase Solutions in $$0

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    Oh sorry for not saying we may not use derivatives yet. – 2017-01-09
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Consider the interval $x\in [2n\pi,(2n+1)\pi]$

$f(2n\pi)=f((2n+1)\pi)=0$

Invoking Rolle's theorem it is clear that there exists atleast one $x=c$ for which $f'(c)=0$ where $c\in(2n\pi,(2n+1)\pi)$.

Also,on this interval $f(x)$ is clearly positive and since $f(x)$ is continuous and differentiable $f(x)$ must achieve atleast one maxima.