Question

Lim(x^2-4)/(x^2-3x+2)

Answer 1

Решаем по правилу лопиталя
$\lim_{x \to \infty} \frac{x^2-4}{x^2-3x+2}\\ \lim_{x \to \infty} \frac{2x}{2x-3}=\lim_{x \to \infty} \frac{2}{2}=1$

Answer 2

To evaluate this limit, we can factor the numerator and denominator and simplify:

lim x->2 (x^2 - 4)/(x^2 - 3x + 2) = lim x->2 [(x + 2)(x - 2)]/[(x - 2)(x - 1)] = lim x->2 (x + 2)/(x - 1)

Now, plugging in x = 2 directly into the above expression would result in division by zero, which is undefined. However, we can use the fact that the limit of a quotient is the quotient of the limits (provided that the denominator does not approach zero), and evaluate the limits of the numerator and denominator separately:

lim x->2 (x + 2)/(x - 1) = lim x->2 (x + 2) / lim x->2 (x - 1) = 4/1 = 4

Therefore, the limit of the original expression is 4.