Question

1) lim (2x*tg4x)/(sin 9x)^2 x->0 2) lim (1+3/x)^2x+1 x->∞

Answer 1

Ответ:

Пределы:

1)

${\boxed{ \lim_{x \to 0} \frac{2x \ \text{tg} \ 4x}{\sin^{2} 9x} = \frac{8}{81} }$

2)

При условии $\boxed{\lim_{x \to \infty} \bigg(1 + \frac{3}{x} \bigg)^{2x + 1} = e^{6}}$

При условии $\boxed{ \lim_{x \to \infty} \Bigg( \bigg(1 + \frac{3}{x} \bigg)^{2x } + 1 \Bigg) = e^{6} + 1}$

Примечание:

Первый замечательный предел:

$\boxed{ \lim_{x \to 0} \frac{\sin x}{x} =1 }$

Следствия из первого замечательного предела:

$\boxed{ \lim_{x \to 0} \frac{\text{tg} \ x}{x} =1 }$

$\displaystyle \lim_{x \to 0} \frac{\text{tg} \ x}{x} = \lim_{x \to 0} \frac{\dfrac{\sin x}{\cos x} }{\dfrac{x}{1} } = \lim_{x \to 0} \frac{\sin x}{x \cos x} = \lim_{x \to 0} \frac{\sin x}{x } \cdot \lim_{x \to 0} \frac{1}{ \cos x}= 1 \cdot \frac{1}{\cos 0} = 1 \cdot 1 =1$

$\boxed{ \lim_{x \to a} \frac{\sin (g(x))}{g(x)} =1 }$ при условии, что $\displaystyle \lim_{x \to a} g(x) = 0$

Второй замечательный предел:

$\boxed{ \lim_{x \to \infty} \bigg(1 + \frac{1}{x} \bigg)^{x} = e}$

Пошаговое объяснение:

1)

$\displaystyle \lim_{x \to 0} \frac{2x \ \text{tg} \ 4x}{\sin^{2} 9x} = \lim_{x \to 0} \frac{2x \cdot 4x \ \text{tg} \ 4x}{4x\sin^{2} 9x} = \lim_{x \to 0} \frac{ \text{tg} \ 4x}{4x} \cdot \lim_{x \to 0} \frac{8x^{2} }{\sin^{2} 9x}=$

$\displaystyle = 1 \cdot 8 \lim_{x \to 0} \frac{x^{2} }{\sin^{2} 9x}= 8 \lim_{x \to 0} \frac{1 }{\dfrac{\sin^{2} 9x}{x^{2} } } = \frac{8}{\displaystyle \lim_{x \to 0} \dfrac{9^{2}\sin^{2} 9x}{9^{2}\cdot x^{2} }} =$

$\displaystyle = \frac{8}{81 \displaystyle \lim_{x \to 0} \dfrac{\sin 9x}{9x } \cdot \lim_{x \to 0} \dfrac{\sin 9x}{9x } } = \dfrac{8}{81 \cdot 1 \cdot 1} = \dfrac{8}{81}$

2)

При условии а):

$\displaystyle \lim_{x \to \infty} \bigg(1 + \frac{3}{x} \bigg)^{2x + 1} = \lim_{x \to \infty} \bigg(1 + \frac{3}{x} \bigg)^{2x} \cdot \lim_{x \to \infty} \bigg(1 + \frac{3}{x} \bigg) =1 \cdot \lim_{x \to \infty} \bigg(1 + \frac{3}{x} \bigg)^{2x}=$

$\displaystyle = \lim_{x \to \infty} \Bigg( \Bigg( \bigg(1 + \frac{1}{\dfrac{x}{3} } \bigg)^{\dfrac{x}{3} } \Bigg)^{3} \Bigg)^{2} = \Bigg( \Bigg( \lim_{x \to \infty} \bigg(1 + \frac{1}{\dfrac{x}{3} } \bigg)^{\dfrac{x}{3} } \Bigg)^{3} \Bigg)^{2} = \bigg((e)^{3} \bigg)^{2} = e^{6}$

При условии б):

$\displaystyle \lim_{x \to \infty} \Bigg( \bigg(1 + \frac{3}{x} \bigg)^{2x } + 1 \Bigg) = \lim_{x \to \infty} (e^{6} + 1) = e^{6} + 1$(часть решения смотрите в пункте а))

Answer 2

To find the limits you mentioned, let's evaluate each of them step by step:

1. lim (2x*tan(4x))/(sin(9x))^2 as x approaches 0:

We can use the fact that sin(x) ~ x for small x and tan(x) ~ x for small x. Therefore, we can rewrite the expression as follows:

lim (2xtan(4x))/(sin(9x))^2 = lim (2x(4x))/(sin(9x))^2 (using the approximations) = lim (8x^2)/(sin(9x))^2

Now, we can use the fact that sin(x) ~ x for small x again to simplify the expression further:

lim (8x^2)/(sin(9x))^2 = lim (8x^2)/(9x)^2 = lim (8/81)

Therefore, the limit is 8/81.

2. lim (1 + 3/x)^(2x+1) as x approaches infinity:

We can rewrite the expression as a limit of the form (1^∞) by using the natural logarithm:

lim (1 + 3/x)^(2x+1) = lim e^(ln((1 + 3/x)^(2x+1))) (applying the natural logarithm)

Now, we can simplify the expression by using the properties of logarithms:

lim e^(ln((1 + 3/x)^(2x+1))) = lim e^((2x+1)*ln(1 + 3/x)) (applying the power rule of logarithms) = e^lim ((2x+1)*ln(1 + 3/x)) (applying the property of e^x being continuous)

Now, let's evaluate the limit inside:

lim ((2x+1)*ln(1 + 3/x)) = lim (2x+1) * lim ln(1 + 3/x) (applying the product rule of limits) = ∞ * ln(1) (since lim (2x+1) = ∞ and lim ln(1 + 3/x) = ln(1) = 0)

Therefore, the limit is ∞.

Please note that these solutions assume that the expressions you provided are correct and there are no errors in the notation or calculation.