Cho hàm số f(x) = căn (1+mx) - căn (1+mx^2)/5x

Ta có \((1 + mx) - \left( {1 + m{x^2}} \right) = \left( {\sqrt {1 + mx}  - \sqrt {1 + m{x^2}} } \right)\left( {\sqrt {1 + mx}  + \sqrt {1 + m{x^2}} } \right)\)

Suy ra \(\sqrt {1 + mx}  - \sqrt {1 + m{x^2}}  = \frac{{mx - m{x^2}}}{{\sqrt {1 + mx}  + \sqrt {1 + m{x^2}} }} = \frac{{mx(1 - x)}}{{\sqrt {1 + mx}  + \sqrt {1 + m{x^2}} }}\)  

Khi đó \(f(x) = \frac{{m(1 - x)}}{{5\left( {\sqrt {1 + mx}  - \sqrt {1 + m{x^2}} } \right)}} \Rightarrow g(x) = \frac{{m(1 - x)}}{{5\left( {\sqrt {1 + mx}  + \sqrt {1 + m{x^2}} } \right)}} \Rightarrow g(0) = \frac{m}{{10}}\)

Vậy giới hạn \(\mathop {\lim }\limits_{x \to 0} f(x) = g(0) = \frac{m}{{10}} = 1 \to m = 10\)