Consider the sequence $$u_n(t)=\sqrt{(t-1)^2 +\frac{1}{n}}.$$ I need to understand if $\|u_n^{\prime}\|_{\infty}\le 1$.
Firstly, $u_n^{\prime}(t)$ stands for the first derivative of $u_n$ made with respect to $t$, isn't it?
If yes, it should be $$u_n^{\prime}(t) =(t-1)\left((t-1)^2 +\frac{1}{n}\right)^{-1/2}.$$
Thus, how to show that $\|u_n^{\prime}\|_{\infty}\le 1$?
Thank you in advance!
Just use the fact that $(t-1)^{2}+\frac 1n \geq (t-1)^{2}$ and $(t-1)^{2})^{-1/2} =\frac 1 {|t-1|}$so $|u_n'(t) |\leq |\frac {t-1} {t-1}|=1$.