Question

Помогите решить пожалуйста, хотя бы несколько

Answer 1

1.
$lim_{n to infty} frac{sin(n^3-4)+2n}{3n^2-1}= lim_{n to infty} frac{2n}{3n^2} = lim_{n to infty} frac{2}{3n} =0$

2.
$lim_{n to infty} frac{1,5n^2+2n-1}{2n^2+n-8} = lim_{n to infty} frac{1,5n^2}{2n^2}=frac{1,5}{2}=0,75$

3.
$lim_{n to infty} frac{ sqrt{3n-1}-sqrt{2n+1} }{sqrt{n}} =lim_{n to infty} frac{ sqrt{3n^2-n}-sqrt{2n^2+n} }{n}= \ lim_{n to infty} frac{ n sqrt{3} -n sqrt{2} }{n} = sqrt{3} - sqrt{2}$

4.
$lim_{x to 1} frac{sin( frac{ pi x}{2} )}{x^2+x-1} = lim_{x to 1} frac{ 1}{1} =1$

5.
$lim_{x to -3} frac{x^2+2x-3}{x^2+x-6}$
x²+2x-3=0
D=2²+4*3=4+12=16    √D=4
x₁=(-2-4)/2=-3
x₂=(-2+4)/2=1
x²+2x-3=(x+3)(x-1)

x²+x-6=0
D=1+4*6=25  √D=5
x₁=(-1-5)/2=-3
x₂=(-1+5)/2=2
x²+x-6=(x+3)(x-2)
$lim_{x to -3} frac{x^2+2x-3}{x^2+x-6} =lim_{x to -3} frac{(x+3)(x-1)}{(x+3)(x-2)}=lim_{x to -3} frac{x-1}{x-2}= frac{-4}{-5} =0,8$

6.
$lim_{x to 0} frac{sin(x^2)}{xtg(2x)}= lim_{x to 0} frac{x^2}{2x^2}= frac{1}{2}$

Answer 2

1.
$lim_{n to infty} frac{sin(n^3-4)+2n}{3n^2-1}=lim_{n to infty} frac{ frac{sin(n^3-4)}{n^2} + frac{2}{n} }{3- frac{1}{n^2} }=0$
2.
$lim_{n to infty} frac{1,5n^2+2n-1}{2n^2+n-8}=lim_{n to infty} frac{1,5+ frac{2}{n} - frac{1}{n^2} }{2+ frac{1}{n} - frac{8}{n^2} }=0,75$
3.
$lim_{n to infty} frac{ sqrt{3n-1}- sqrt{2n+1}}{sqrt n}=lim_{n to infty} frac{3n-1-(2n+1)}{sqrt{n}(sqrt{3n-1}+sqrt{2n+1})}=$$lim_{n to infty} frac{n-2}{n(sqrt{3- frac{1}{n} }+sqrt{2+frac{1}{n}})}=lim_{n to infty} frac{1- frac{2}{n} }{sqrt{3- frac{1}{n} }+sqrt{2+frac{1}{n}}}=frac{1}{sqrt3+sqrt2}=sqrt3-sqrt2$
4.
$lim_{x to1} frac{sin( frac{ pi x}{2})}{x^2+x-1}=frac{sin frac{ pi}{2}}{1^2+1-1}=sin frac{ pi}{2}}=1$
5.
$lim_{x to-3} frac{x^2+2x-3}{x^2+x-6}=lim_{x to-3} frac{(x-1)(x+3)}{(x-2)(x+3)}=lim_{x to-3} frac{x-1}{x-2}=frac{-4}{-5}=frac{4}{5}$
6.
$lim_{x to0} frac{sin(x^2)}{xtg(2x)}=lim_{x to0} frac{x^2}{x*2x}=lim_{x to0} frac{x^2}{2x^2}=frac{1}{2}$
7.
$lim_{x to infty}( frac{x^2+2}{x^2+1})^{x^2+3x-1}=lim_{x to infty}( frac{(x^2+1)+1}{x^2+1})^{(x^2+1)+(3x-2)}=$$lim_{x to infty}[(1+frac{1}{x^2+1})^{x^2+1}*(1+frac{1}{x^2+1})^{3x-2}]=$$lim_{x to infty}(1+frac{1}{x^2+1})^{x^2+1}*lim_{x to infty}(1+frac{1}{x^2+1})^{3x-2}=$$e*lim_{x to infty}(1+frac{1}{x^2+1})^{3x-2}$
7-й пока не выходит...

Answer 3

В седьмом будет e

Answer 4

Думаю, вы правы. Но как корректно это показать?