Question

[tex] \lim_{x[tex] \lim_{x \to 4\ \frac{\sqrt{x}-2 }{x-4} Решить лимиты.

Answer 1

$\lim_{x \to \ 4} \frac{ \sqrt{x} -2}{x-4} = \frac{ \sqrt{4}-2 }{4-4} = \frac{0}{0}$
неопределенность 0/0
$\lim_{x \to \ 4} \frac{ \sqrt{x} -2}{x-4} = \lim_{x \to \ 4} \frac{ \sqrt{x} -2}{ ( \sqrt{x} )^{2} - 2^{2} } = \lim_{x \to \ 4} \frac{ \sqrt{x} -2}{( \sqrt{x} -2)*( \sqrt{x} +2)} =$
$= \lim_{x \to \ 4} \frac{1}{ \sqrt{x} +2} = \frac{1}{ \sqrt{4}+2 } = \frac{1}{4}=0,25$

Answer 2

To solve this limit, we can try to simplify the expression using algebraic manipulations. First, we notice that the expression has a factor of $\sqrt{x}-2$ in the numerator and a factor of $x-4$ in the denominator. We can use the identity $a^2-b^2=(a-b)(a+b)$ to simplify the numerator:

\begin{align*} \sqrt{x}-2 &= (\sqrt{x}-2)\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} \ &= \frac{x-4}{\sqrt{x}+2} \end{align*}

Substituting this into the original expression gives:

\begin{align*} \lim_{x\to 4} \frac{\sqrt{x}-2}{x-4} &= \lim_{x\to 4} \frac{\frac{x-4}{\sqrt{x}+2}}{x-4} \ &= \lim_{x\to 4} \frac{1}{\sqrt{x}+2} \ &= \frac{1}{\sqrt{4}+2} \ &= \frac{1}{4} \end{align*}

Therefore, the limit is $\frac{1}{4}$.