Question

Lim x->2/3 3x^3+5x+2/3x+8x+4

Answer 1

Ответ:

10

Объяснение:

$\lim_{x \to 2/3} 3x^3+5x+2/3x+8x+4$

Внутри предела - многочлен, а по свойству предела $\lim_{n \to a} P(n)=P(a)$, где Р(n) - многочлен.

Тогда $\lim_{x \to 2/3} 3x^3+5x+2/3x+8x+4=3(2/3)^3+5(2/3)+2/3(2/3)+8(2/3)+4=8/9+10/3+4/9+16/3=8/9+30/9+4/9+48/9=\frac{8+30+4+48}{9} =\frac{90}{9} =10$

Answer 2

It looks like you're asking about the limit of a rational function as x approaches 2/3. The given expression is:

$\lim_{{x \to \frac{2}{3}}} \frac{3x^3 + 5x + 2}{3x + 8x + 4}$

To evaluate this limit, you can factor out common terms in the numerator and denominator:

$\lim_{{x \to \frac{2}{3}}} \frac{x^3(3 + \frac{5}{x^2} + \frac{2}{x^3})}{x(3 + 8 + \frac{4}{x})}$

As $x$ approaches $\frac{2}{3}$, all the terms with $x$ in the denominator will approach 0. So, you're left with:

$\frac{\frac{2}{3}^3 \cdot (3 + \frac{5}{{\frac{2}{3}}^2} + \frac{2}{{\frac{2}{3}}^3})}{1 \cdot (3 + 8 + \frac{4}{\frac{2}{3}})}$

Simplify the terms in the numerator:

$\frac{\frac{8}{27} \cdot (3 + \frac{5}{\frac{4}{9}} + \frac{2}{\frac{8}{27}})}{1 \cdot (3 + 8 + \frac{18}{2})}$

$\frac{\frac{8}{27} \cdot (3 + \frac{45}{4} + \frac{54}{8})}{1 \cdot (3 + 8 + 9)}$

$\frac{\frac{8}{27} \cdot (\frac{12 + 45 + 54}{4})}{1 \cdot 20}$

$\frac{\frac{8}{27} \cdot \frac{111}{4}}{20}$

$\frac{\frac{2968}{108}}{20}$

$\frac{173}{135}$

So, the limit of the given expression as $x$ approaches $\frac{2}{3}$ is $\frac{173}{135}$.