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

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