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
 A: hint: $a-b = \frac{a^2-b^2}{a+b}$
A: 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.
A: This is probably not an orthodox approach to limit problems, unlike the solutions given above by others, but here it is anyway:
$$\begin{align}
\require{cancel}
&\lim_{x\to\infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2-x}\right)\\
&=\lim_{x\to\infty}\left[ x\left(1+\frac 1x+\frac1{x^2}\right)^{1/2}-x\left(1-\frac1x\right)^{1/2}\right]\\
&=\lim_{x\to\infty}
\bigg\lbrace x
         \left[
             1+\frac 12\left(\frac 1x+\frac1{x^2}\right)
              +\frac{\frac 12\left(-\frac 12\right)}{1\cdot 2}\left(\frac 1x+\frac1{x^2}\right)^2+\cdots \right]
        - x
          \left[
              1+\frac 12\left(-\frac1x\right)
               +\frac{\frac12\left(-\frac12\right)}{1\cdot 2}\left(-\frac 1x\right)^2+\cdots \right]
              \bigg\rbrace\\
&=\lim_{x\to\infty}
\bigg\lbrace x
         \left[
             \cancel{1}+\frac 1{2x}+\frac3{8x^2}\cdots \right]
        - x
          \left[\cancel{1}-\frac 1{2x}-\frac1{8x^2}+\cdots \right]
       \bigg\rbrace\\
&=\lim_{x\to\infty}
\left[1+\frac1{2x}+\cdots\right]\\
&=1\qquad \blacksquare
              \end{align}$$
A: 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 $1$.
