Question

Lim (x->o) x^(sin x)

Answer 1

Логарифмируем предел и получаем:

$\lim_{x \to 0}\ln(x^{\sin x})= \lim_{x \to 0}{\sin x}*\ln x=\lim_{x \to 0}x*\ln x\\=\lim_{x \to 0}\frac{\ln x}{\frac{1}{x} } =\lim_{x \to 0}\frac{1/x}{-\frac{1}{x^2} }= -\lim_{x \to 0} x=0/$

Потенцируем и находим окончательный ответ

$e^0=1/$

Answer 2

To evaluate the limit $\lim_{{x \to 0}} x^{\sin x}$, we need to consider the behavior of the expression as $x$ approaches 0.

When dealing with limits of the form $0^0$, $1^{\infty}$, or $\infty^0$, we often encounter indeterminate forms, and the limit could be anything. To handle this limit, we can use L'Hôpital's rule or rewrite the expression using some algebraic manipulations and apply known limits.

Let's use the natural logarithm to rewrite the expression:

$y = \lim_{{x \to 0}} x^{\sin x}$

Taking the natural logarithm of both sides:

$\ln y = \lim_{{x \to 0}} \ln(x^{\sin x})$

Using the logarithm property $\ln(a^b) = b \ln(a)$:

$\ln y = \lim_{{x \to 0}} \sin x \ln x$

Now, we can consider the limit of $\ln y$ instead, as this may make it easier to work with:

$\lim_{{x \to 0}} \ln y = \lim_{{x \to 0}} \sin x \ln x$

Now, we know that $\lim_{{x \to 0}} \ln x = -\infty$ (the natural logarithm of 0 is negative infinity). Also, $\lim_{{x \to 0}} \sin x = 0$. Therefore, we have an indeterminate form of $\infty \times 0$.

To resolve this indeterminate form, we can apply L'Hôpital's rule, which states that for the indeterminate form $\infty \times 0$ (and some other indeterminate forms), the limit of the ratio of derivatives of the numerator and denominator will give us the same limit. So, taking the derivatives:

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

Now we have another indeterminate form, $0 / 0$, so we apply L'Hôpital's rule again:

$\lim_{{x \to 0}} \frac{\ln x}{\frac{1}{\sin x}} = \lim_{{x \to 0}} \frac{\frac{1}{x}}{\frac{-1}{\sin^2 x}}$

Simplifying:

$\lim_{{x \to 0}} \frac{\frac{1}{x}}{\frac{-1}{\sin^2 x}} = \lim_{{x \to 0}} \frac{-\sin^2 x}{x} = 0$

So, $\lim_{{x \to 0}} \ln y = 0$. Now we can find the value of $y$ by exponentiating both sides:

$y = e^{\lim_{{x \to 0}} \ln y} = e^0 = 1$

Thus, $\lim_{{x \to 0}} x^{\sin x} = 1$.