Теория:

Первый замечательный предел:

[tex]\displaystyle \lim_{x \to 0} \frac{sin(x)}{x}=1[/tex]

Следствие:

[tex]\displaystyle \lim_{x \to 0} \frac{tg(x)}{x}=1[/tex]

Решение:

1) [tex]\displaystyle \lim_{x \to 0} \frac{sin^2(5x)}{25x^2}= \lim_{x \to 0} \frac{sin(5x)*sin(5x)}{5x*5x} = \lim_{x \to 0} (\frac{sin(5x)}{5x}*\frac{sin(5x)}{5x})=1*1=1[/tex]

2) [tex]\displaystyle \lim_{x \to 0} \frac{sin^2(6x)}{x^2}= \lim_{x \to 0} \frac{sin(6x)*sin(6x)}{x*x} *\frac{6*6}{6*6} = \lim_{x \to 0} \frac{36*sin(6x)*sin(6x)}{6x*6x} =\lim_{x \to 0} (36*\frac{sin(6x)}{6x} *\frac{sin(6x)}{6x} ) = 36*1*1 = 36[/tex]

3) [tex]\displaystyle \lim_{x \to 0} \frac{tg^2(2x)}{4x^2} =\lim_{x \to 0} \frac{tg(2x)*tg(2x)}{2x*2x}=\lim_{x \to 0} (\frac{tg(2x)}{2x}*\frac{tg*(2x)}{2x})=1*1=1[/tex]