Question

Lim x → 0 (6-5/cosx)^ctg^2(x^2)

Answer 1

Если подставить х=0, то получим неопределенность $\{1^{\infty}\}$. Тогда можем применить второй замечательный предел.

$\displaystyle \lim_{x \to 0} \bigg(6-\frac{5}{\cos x}\bigg)^{ctg^2x^2}=\lim_{x \to 0} \bigg(1+\frac{5\cos x-5}{\cos x}\bigg)^{ctg^2x^2\cdot\frac{5\cos x-5}{\cos x}\cdot\frac{\cos x}{5\cos x-5}} =\\ \\ \\ =e^\big{\lim_{x \to 0} \frac{5(\cos x-1)}{tg^2x^2\cos x}} =e^\big{\lim_{x \to 0} \frac{-10\sin^2\frac{x}{2}}{x^4}} =e^\big{\lim_{x \to 0}\frac{-10\cdot\frac{x^2}{4}}{x^4} } =e^{-\infty}=0$

Answer 2

To evaluate the given limit, let's break it down step by step. First, let's simplify the expression inside the parentheses:

1. Simplify the expression inside the parentheses: (6 - 5/cos(x))

To simplify this, we need to find the common denominator for 6 and 5/cos(x), which is cos(x). So we get:

(6*cos(x) - 5)/cos(x)

Now, let's proceed to the next step:

2. Simplify the exponent term: ctg^2(x^2)

The cotangent function, cot(x), is the reciprocal of the tangent function, 1/tan(x). The square of cot(x) is cot^2(x) = (1/tan(x))^2 = 1/tan^2(x). And since cot(x) = cos(x)/sin(x), we can rewrite cot^2(x) as:

cot^2(x) = (cos^2(x)/sin^2(x))

Now, since x → 0, x^2 → 0 as well, and sin(0) = 0. So the denominator sin^2(x) will be 0. Hence, the entire expression ctg^2(x^2) will become undefined at x = 0.

As a result, the limit of the given expression as x approaches 0 cannot be directly evaluated since the exponent term becomes undefined.

It's possible that there might be an error or misunderstanding in the original expression provided. If you have any clarifications or additional context, please feel free to share them, and I'd be happy to assist further.