Вычислить пределы функций

По определению:
$\lim_{x \to a} f(x) = f(a) \\$
a)
$\lim_{x \to 1} \frac{\sqrt{x}-1}{x^2-1} = \lim_{x \to 1} \frac{\sqrt{x}-1}{(x-1)(x+1)} = \lim_{x \to 1} \frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)} = \\\\ = \lim_{x \to 1} \frac{1}{(\sqrt{x}+1)(x+1)} = \frac{1}{(1+1)(1+1)} = \frac{1}{4}\\$

b)
$\lim_{x \to \infty} \frac{2x^2-3}{4x^3+5x} = \lim_{x \to \infty} \frac{x^3(\frac{2}{x}-\frac{3}{x^3})}{x^3(4+\frac{5}{x^2})} = \lim_{x \to \infty} \frac{\frac{2}{x}-\frac{3}{x^3}}{4+\frac{5}{x^2}} = \frac{0-0}{4+0} = \frac{0}{4} = 0\\$

v)
$\lim_{x \to 0} tg2xctg4x = \lim_{x \to 0} tg2x\frac{ctg^22x-1}{2ctg2x} = \lim_{x \to 0} \frac{tg2xctg^22x-tg2x}{2ctg2x} = \\\\ \lim_{x \to 0}\frac{ctg2x-tg2x}{2ctg2x} = \lim_{x \to 0} \frac{ctg2x}{2ctg2x}-\frac{tg2x}{2ctg2x}=\lim_{x \to 0} \frac{1}{2}-\frac{tg^22x}{2} = \\\\ =\frac{1}{2} - \frac{0}{2} = \frac{1}{2}$