My solution is as follow.
Can anyone tell if my calculation is correct.
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makes sense to me, but i don't understand what it has to do with triangle inequality, it is obvious that $\left|x^2+1\right| \ge \left| x^2 \right|$ – gt6989b May 27 '22 at 03:56
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did you mean I will take x^2<epsilon? – William Bani May 27 '22 at 04:02
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Writing $\left|\frac x{x^2+1}\right|\leq\frac 1{|x|}$ does not help. You need to write it as $\left|\frac x{x^2+1}\right|\leq |x|$, using $x^2+1\geq 1$. – Pythagoras May 27 '22 at 04:05