Multiply by the conjugate and then simplify:
x 2 + 2 x − x 2 − 2 x = ( x 2 + 2 x − x 2 − 2 x ) ⋅ x 2 + 2 x + x 2 − 2 x x 2 + 2 x + x 2 − 2 x = x 2 + 2 x − ( x 2 − 2 x ) x 2 + 2 x + x 2 − 2 x = 4 x x 2 ( 1 + 2 / x ) + x 2 ( 1 − 2 / x ) = 4 x x ( 1 + 2 / x + 1 − 2 / x ) = 4 1 + 2 / x + 1 − 2 / x {\displaystyle {\begin{aligned}{\sqrt {x^{2}+2x}}-{\sqrt {x^{2}-2x}}&=({\sqrt {x^{2}+2x}}-{\sqrt {x^{2}-2x}})\cdot {\frac {{\sqrt {x^{2}+2x}}+{\sqrt {x^{2}-2x}}}{{\sqrt {x^{2}+2x}}+{\sqrt {x^{2}-2x}}}}\\&={\frac {x^{2}+2x-(x^{2}-2x)}{{\sqrt {x^{2}+2x}}+{\sqrt {x^{2}-2x}}}}\\&={\frac {4x}{{\sqrt {x^{2}(1+2/x)}}+{\sqrt {x^{2}(1-2/x)}}}}\\&={\frac {4x}{x({\sqrt {1+2/x}}+{\sqrt {1-2/x}})}}\\&={\frac {4}{{\sqrt {1+2/x}}+{\sqrt {1-2/x}}}}\end{aligned}}}
assuming x > 0 {\displaystyle x>0} . Hence
lim x → + ∞ ( x 2 + 2 x − x 2 − 2 x ) = lim x → + ∞ 4 1 + 2 / x + 1 − 2 / x = 4 2 = 2 {\displaystyle {\begin{aligned}\lim _{x\rightarrow +\infty }({\sqrt {x^{2}+2x}}-{\sqrt {x^{2}-2x}})&=\lim _{x\rightarrow +\infty }{\frac {4}{{\sqrt {1+2/x}}+{\sqrt {1-2/x}}}}\\&={\frac {4}{2}}=2\end{aligned}}}
. Hence
