Question

Lim(x->0 y->0) (1+x^2+y^2)^(1/(x^2+y^2))=?

Answer 1

Ответ:

$\displaystyle \lim_{x \to 0} \bigg ( \lim_{y \to 0} (1+x^2+y^2)^{\displaystyle\frac{1}{1+x^2+y^2}}\bigg )=e$

Пошаговое объяснение:

если это повторный предел, то тогда так

$\displaystyle \lim_{x \to 0} \bigg ( \lim_{y \to 0} (1+x^2+y^2)^{\displaystyle\frac{1}{1+x^2+y^2}}\bigg )$

посчитаем внутренний предел

$\displaystyle \lim_{y \to 0} (1+x^2+y^2)^{\displaystyle\frac{1}{1+x^2+y^2}}=(1+x^2)^{\displaystyle\frac{1}{x^2} }$

теперь посчитаем внешний предел

для этого используем свойства второго замечательного предела

$\displaystyle \lim_{x \to 0}\bigg (1+\frac{a}{x} \bigg )^\displaystyle{bx}}= e^\displaystyle{ab}$

$\displaystyle \lim_{x \to 0} (1+x^2)^\displaystyle{\frac{1}{x^2} } }=\bigg [a=x^3 \quad b=\frac{1}{x^3} \bigg ]=e^{\displaystyle x^3*\frac{1}{x^3} }}=e$

ответ

$\displaystyle \lim_{x \to 0} \bigg ( \lim_{y \to 0} (1+x^2+y^2)^{\displaystyle\frac{1}{1+x^2+y^2}}\bigg )=e$

Answer 2

To find the limit of the expression as (x, y) approaches (0, 0), you can use L'Hôpital's Rule or another approach. Let's use L'Hôpital's Rule:

First, rewrite the limit as follows:

lim (x->0, y->0) (1 + x^2 + y^2)^(1/(x^2 + y^2))

Now, let's take the natural logarithm (ln) of both sides:

ln[lim (x->0, y->0) (1 + x^2 + y^2)^(1/(x^2 + y^2))]

Using the properties of the natural logarithm, you can bring the exponent down:

lim (x->0, y->0) ln[(1 + x^2 + y^2)^(1/(x^2 + y^2))]

Now, the limit of the natural logarithm is easier to handle. Apply it to the expression inside the ln:

ln[lim (x->0, y->0) (1 + x^2 + y^2)^(1/(x^2 + y^2))]

= ln[lim (x->0, y->0) (1 + x^2 + y^2)^(1/(x^2 + y^2))]

= ln[1]

Since the natural logarithm of 1 is 0, the limit of the original expression is:

e^0 = 1

So, lim (x->0, y->0) (1 + x^2 + y^2)^(1/(x^2 + y^2)) = 1.