Question

Lim tg x/x = 1 xстремиться к 0 и lim x/tgx = 1 xстремиться к 0 одно и тоже?

Answer 1

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Answer 2

Lim tg(x)/x = 1 and lim x/tg(x) = 1: Are they the same?

The limits lim tg(x)/x and lim x/tg(x) both approach 1 as x approaches 0. Therefore, they are indeed the same.

To understand why these limits are equal, let's examine each limit separately:

1. lim tg(x)/x as x approaches 0: - The limit of tg(x)/x as x approaches 0 can be evaluated using L'Hôpital's rule, which states that if the limit of the ratio of two functions is of the form 0/0 or ∞/∞, then the limit of their derivatives is the same as the original limit. - Taking the derivative of the numerator and denominator, we get lim (sec^2(x))/1 as x approaches 0. - The derivative of sec^2(x) is 2sec(x)tan(x). - Substituting x = 0 into the derivative, we get 2sec(0)tan(0) = 0. - Therefore, lim tg(x)/x as x approaches 0 is equal to lim (sec^2(x))/1 as x approaches 0, which is equal to 0/1 = 0.

2. lim x/tg(x) as x approaches 0: - The limit of x/tg(x) as x approaches 0 can also be evaluated using L'Hôpital's rule. - Taking the derivative of the numerator and denominator, we get lim (1)/(sec^2(x)) as x approaches 0. - The derivative of sec^2(x) is 2sec(x)tan(x). - Substituting x = 0 into the derivative, we get 1/(sec^2(0)) = 1/1 = 1. - Therefore, lim x/tg(x) as x approaches 0 is equal to lim (1)/(sec^2(x)) as x approaches 0, which is equal to 1/1 = 1.

Hence, both limits lim tg(x)/x and lim x/tg(x) approach 1 as x approaches 0, indicating that they are indeed the same.