Помогите решить, завтра работу нужно сдать(( в баллах не обижу

Question 1

Question 2

1) f(x) = -5x^2 - 8; x0 = 2
f(2) = -5*2^2 - 8 = -5*4 - 8 = -28
lim(x->2) (-5x^2 - 8) = -5*2^2 - 8 = -5*4 - 8 = -28

2)
$\lim_{n \to \infty} \frac{4n-3}{2n+1} = \lim_{n \to \infty} \frac{4-3/n}{2+1/n}= \frac{4-0}{2+0}=2$

3)
$\lim_{x \to 2} \frac{3x^2-5x-2}{x-2}= \lim_{x \to 2} \frac{(x-2)(3x+1)}{x-2} =\lim_{x \to 2} (3x+1)=7$

4)
$\lim_{x \to 1} \frac{x^2-2x+1}{2x^2-x-1}= \lim_{x \to 1} \frac{2x-2}{4x-1} = \frac{2-2}{4-1} =0$

5)
$\lim_{x \to 0} \frac{arcsin(3x)}{ \sqrt{2+x}- \sqrt{2}} = \lim_{x \to 0} ( \frac{arcsin(3x)}{3x} *3x* \frac{\sqrt{2+x}+ \sqrt{2}}{(\sqrt{2+x}- \sqrt{2})(\sqrt{2+x}+ \sqrt{2})} ) =$
$= \lim_{x \to 0} (1*3x* \frac{\sqrt{2+x}+ \sqrt{2}}{2+x-2} ) = \lim_{x \to 0} (3(\sqrt{2+x}+ \sqrt{2})) =6 \sqrt{2}$

6)
$\lim_{x \to 1} \frac{ \sqrt{x^2-x+1}-1 }{tg(pi*x)} = \lim_{x \to 1} ( \frac{(\sqrt{x^2-x+1}-1)(\sqrt{x^2-x+1}+1)}{\sqrt{x^2-x+1}+1} * \frac{pi*x}{tg(pi*x)}* \frac{1}{pi*x} )$
$\lim_{x \to 1} \frac{(x^2-x+1-1)}{\sqrt{x^2-x+1}+1} * \lim_{x-1 \to 0} (\frac{pi*x-pi}{tg(pi*x-pi)}* \frac{1}{pi*x-pi} )=$
$\lim_{x \to 1} \frac{(x^2-x)}{\sqrt{x^2-x+1}+1} * \lim_{x-1 \to 0} (1* \frac{1}{pi*x-pi} )=$
$=\lim_{x \to 1} (\frac{x(x-1)}{\sqrt{x^2-x+1}+1} * \frac{1}{pi(x-1)})= \frac{1}{(1+1)*pi} = \frac{1}{2pi}$