<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] Multiply numerator and denominator by the conjugate of the "numerator": $\sqrt{x^2 + x+1} + \sqrt{x^2 -x}$, to get a difference of squares in the numerator.
$$\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)\cdot \frac{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}$$ 
$$ = \lim_{x\to \infty} \frac{x^2 + x + 1 -(x^2 -x)}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}
$$ $$ = \lim_{x\to \infty} \frac{2x+1}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}$$
Now, divide numerator and denominator by $x$:
$$ = \lim_{x\to \infty} \frac{2+\frac 1x}{\sqrt{1+\frac 1x +\frac 1{x^2}} + \sqrt{1 -\frac{1}{x}}}= \frac 22 = 1$$

So whenever you encounter limits of the form $$\lim_{x\to \infty} \left(\sqrt{f(x)} - \sqrt{g(x)}\right) \leadsto \infty - \infty,$$ "multiply by $1$". That is, multiply  by $\dfrac{\sqrt{f(x)} + \sqrt{g(x)}}{\sqrt{f(x)} + \sqrt{g(x)}}.\;$ You'll likely be using this strategy in a whole host of situations.
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