Regarding the supremum of the given function

Question

Given $n\in\mathbb{N}$. Can one help me with the

$$\sup\left\{\frac{t}{\sqrt{1+n^2t^2}} : t\geq 0\right\}?$$

Will it be less than $\frac{1}{n}$? I tried differentiating and equating to zero. But it is getting complicated.

Answer 1

If $t\neq 0$, you can rewrite the expression as

$$\frac{t}{\sqrt{1+n^2t^2}} = \frac{\frac1t}{\frac1{\sqrt{t^2}}}\cdot\frac{t}{\sqrt{1+n^2t^2}} = \frac{\frac1t\cdot t}{\frac{\sqrt{1+n^2t^2}}{\sqrt{t^2}}} =\frac{1}{\sqrt{\frac{1}{t^2} + n^2}}$$

Now, note that:

• $\frac{1}{t^2}$ is a decreasing function.
• Therefore, $\frac{1}{t^2} + n^2$ is also decreasing.
• Therefore, $\sqrt{\frac{1}{t^2} + n^2}$ is also decreasing.
• Therefore, $\frac{1}{\sqrt{\frac{1}{t^2} + n^2}}$ is increasing.

Answer 2

Let $f(t)=\frac{t}{\sqrt{1+n^2t^2}}$ for $t \ge 0.$ Then $f(0)=0$ and $f(t) >0$ for $t>0.$ $f$ is differentiable.

Show that $f'(t) >0$ for $t \ge 0.$ Hence $f$ is strictly increasing in $[0, \infty)$.

Conclusion:

$$\sup\left\{\frac{t}{\sqrt{1+n^2t^2}} : t\geq 0\right\}= \lim_{t \to \infty}f(t).$$

Answer 3

Consider $$\Delta=\frac{t}{\sqrt{1+n^2 t^2}}-\frac{1}{n}$$ and expand it for large values of $t$ to obtain $$\Delta=-\frac{1}{2 n^3 t^2}\left(1-\frac{3}{4 n^2 t^2}+O\left(\frac{1}{t^4}\right)\right)$$ and since the derivative does not cancel, then $\cdots$