<s>[INST] Find $\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)$ I am having a tough time with these TYPE of problems
looking forward  ideas, All ideas will be appreciated
 [/INST] Here is another approach:
$\sqrt{x^2+x+1} - \sqrt{x^2-x} = x (\sqrt{1+{1 \over x}+{1 \over x^2}} - \sqrt{1-{1 \over x}} )$.
For $\delta$ sufficiently small, we have $|\sqrt{1 + \delta}-(1+{\delta \over 2}) | \le |\delta|^2 $ (this follows from the Taylor expansion of $\delta \mapsto \sqrt{1+\delta}$, which is $1+{1 \over 2} \delta - {1 \over 8} \delta^2 + \cdots$).
Hence we have
$|\sqrt{1+{1 \over x}+{1 \over x^2}} - (1+ {1 \over 2}({1 \over x}+{1 \over x^2}))| \le |{1 \over x}+{1 \over x^2}|^2 $ and
$|\sqrt{1-{1 \over x}}-(1- {1 \over 2} {1 \over x})| \le | {1 \over x} |^2 $.
Consequently, we have
$|\sqrt{1+{1 \over x}+{1 \over x^2}} -\sqrt{1-{1 \over x}}-({1 \over x}+{1 \over 2} {1 \over x^2})| \le  |{1 \over x}+{1 \over x^2}|^2 + | {1 \over x} |^2  $.
Now multiply by $x$ to get
$|x(\sqrt{1+{1 \over x}+{1 \over x^2}} -\sqrt{1-{1 \over x}})-(1+{1 \over 2} {1 \over x})| \le  x(|{1 \over x}+{1 \over x^2}|^2 + | {1 \over x} |^2 ) $.
Now let $x \to \infty$, and we see that the limit is $1$.
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