Lim x→-1 5x^2+4x-1\3x^2+x-2

Предыдущий
вопрос Следующий
вопрос

To find the limit as x approaches -1 of the expression (5x^2 + 4x - 1) / (3x^2 + x - 2), we can plug in x = -1 and simplify:

(5(-1)^2 + 4(-1) - 1) / (3(-1)^2 + (-1) - 2)

= (5(1) - 4 - 1) / (3(1) - 1 - 2)

= (5 - 4 - 1) / (3 - 1 - 2)

= 0 / 0

Since we got an indeterminate form of 0/0, we can apply L'Hopital's Rule which states that if the limit of a quotient of two functions is of the form 0/0 or ∞/∞, then we can take the derivative of the numerator and denominator separately and then evaluate the limit.

Taking the derivative of the numerator and denominator, we get:

d/dx (5x^2 + 4x - 1) = 10x + 4

d/dx (3x^2 + x - 2) = 6x + 1

Now, we can rewrite the expression in terms of these derivatives:

lim x→-1 (5x^2 + 4x - 1)/(3x^2 + x - 2) = lim x→-1 (10x + 4)/(6x + 1)

Plugging in x = -1:

= (10(-1) + 4) / (6(-1) + 1)

= (-10 + 4) / (-6 + 1)

= -6 / -5

= 6/5

Therefore, lim x→-1 (5x^2 + 4x - 1)/(3x^2 + x - 2) = 6/5.