Question

Limx⇒1 (√x-1)/(∛x²-1)

Answer 1

При подставке мы получаем неопределенность 0/0. Соотвественно нужно взять по отдельности производные от числителя и знаменателя

Answer 2

To find the limit of the given expression as x approaches 1, you can try to simplify the expression and then evaluate the limit.

Start by rationalizing the numerator and denominator to simplify the expression:

√x - 1 = (√x - 1) * (√x + 1) / (√x + 1) = (x - 1) / (√x + 1)

∛x² - 1 = (∛x² - 1) * (∛x⁴ + ∛x² + 1) / (∛x⁴ + ∛x² + 1) = (x² - 1) / (∛x⁴ + ∛x² + 1)

Now, the original expression can be rewritten as:

lim (x → 1) [(x - 1) / (√x + 1)] / [(x² - 1) / (∛x⁴ + ∛x² + 1)]

Next, let's simplify the expression further:

= lim (x → 1) [(x - 1) / (√x + 1)] * [((∛x⁴ + ∛x² + 1) / (x² - 1))]

Now, you can evaluate the limit. Plug in x = 1:

= [(1 - 1) / (√1 + 1)] * [((∛1⁴ + ∛1² + 1) / (1² - 1))]

= (0 / 2) * [(1 + 1 + 1) / (1 - 1)]

= 0 * (3 / 0)

This is an indeterminate form (0 times infinity). To evaluate it further, you can apply L'Hôpital's Rule. Take the derivative of the numerator and denominator and then evaluate the limit again:

The derivative of the numerator is 0, and the derivative of the denominator is also 0. So, you have an indeterminate form (0/0).

Apply L'Hôpital's Rule again:

lim (x → 1) [0' / 0'] / [0' / 0']

Now, the derivatives of both the numerator and denominator are still 0, resulting in another indeterminate form (0/0).

You can continue applying L'Hôpital's Rule until you get a determinate form or simplify the expression differently, but it seems that this limit is approaching an indeterminate form that may require more advanced techniques to evaluate.