Question

Найти указанные пределы, не пользуясь правилом Лопиталя.

Answer 1

А) 4x²+7x+3=(4x+3)•(x+1)
2x²+x-1=(2x-1)•(x+1)
При х→-1
=lim[(4x+3)/(2x-1)]=-1/(-3)=⅓.

б) числитель и знаменатель делишь на х²:
=lim[(3/x²-2/x-1)/(1+4/x+1/x²)]=-1/1=-1.

в) заменяешь на эквивалентные бесконечно малые.
При х→0
arcsin2x ~ 2x
=lim(2x/4x)=½.

г) при х→∞
lim[(4x+3)/(4x-1)]^(2x+3)=lim[1+4/(4x-1)]^[((4x-1)/4)•(4/(4x-1))•(2x+3)=
=e^lim[(8x+12)/(4x-1)]=e². 
УДАЧИ:)

Answer 2

Finding Limits Without Using L'Hôpital's Rule

To find the limits without using L'Hôpital's Rule, we can use other methods such as factoring, rationalizing, or using trigonometric identities.

The given equation is x^2 - 6x + 8 = 0. The discriminant D is calculated as D = (-6)^2 - 4*1*8 = 4.

Calculating the Limit

To calculate the limit, we can use various techniques such as substitution, factoring, or trigonometric identities, depending on the specific limit problem.

Without the specific limit to calculate, it's challenging to provide a detailed solution. If you have a specific limit in mind, please provide the function and the point at which the limit is to be evaluated, and I can assist you further.

If you have a specific example or problem in mind, feel free to share it, and I can help you work through the process of finding the limit without using L'Hôpital's Rule.