1)lim 1-cosx/x²
x→0
2) lim arctg7x(e^2x -1)
x→0
3) lim 2+∛x+2/x+10
x→-10
4) lim x²-5x+6/x²-12x+20
x→2

Предыдущий
вопрос Следующий
вопрос

1) To find the limit of (1-cosx)/x^2 as x approaches 0, we can use L'Hopital's Rule. Taking the derivative of the numerator and denominator gives sin(x) in the numerator and 2x in the denominator. So, the limit becomes lim x->0 sin(x)/2x. As sin(x)/x approaches 1 as x approaches 0, the limit of the original function is 1/2.

2) To find the limit of arctan(7x)(e^(2x) - 1) as x approaches 0, we can substitute x=0 into the expression, giving arctan(0)(e^0 - 1) = 0.

3) To find the limit of (2 + cuberoot(x+2))/(x+10) as x approaches -10, we can substitute x=-10 into the expression, giving (2 + cuberoot(-8))/0. This is undefined as division by zero is not allowed.

4) To find the limit of (x^2-5x+6)/(x^2-12x+20) as x approaches 2, we can factorize the numerator and denominator to get (x-2)(x-3)/(x-2)(x-10). Canceling out the common factor (x-2), we get the limit as 1/(2-10) = -1/8.