Question

resolva a questão abaixo a respeito dos Limites de uma função

Answer 1

a+b+c é igual a 12

Para que f seja contínua em x = a, os limites laterais no ponto x = a devem existir e ser iguais.

$\displaystyle i) \lim_{x \to 1-} \dfrac{|x^2-5x+4|}{x-1} = \lim_{x \to 1-} \dfrac{|x-4|.\overbrace{|x-1|}^{<0}}{x-1} = \lim_{x \to 1-} -|x-4| = 3$

$\displaystyle ii) \lim_{x \to 1+} ax+b = 3 \Rightarrow \boxed{\boxed{a + b = 3}}$

$\textsf{--------------------}$$\textsf{--------------------}$$\textsf{--------------------}$$\textsf{--------------------}$

$\displaystyle iii) \lim_{x \to 2+}\bigg\{(x-3)\cdot \left[\dfrac{\sqrt{x} - \sqrt{2} + \sqrt{x-2}}{\sqrt{x^2-4}}\right] + 9\bigg\} \Longleftrightarrow \\\\\\\lim_{x \to 2+}\bigg\{ (x-3)\cdot \left[\dfrac{\sqrt{x-2}}{\sqrt{x+2} \cdot (\sqrt{x}+\sqrt{2}) }+\dfrac{1}{\sqrt{x+2}}\right] + 9\bigg\} \Longleftrightarrow \\\\\\\ \bigg\{ (2-3)\cdot \left[\dfrac{\sqrt{2-2}}{\sqrt{2+2} \cdot (\sqrt{2}+\sqrt{2}) }+\dfrac{1}{\sqrt{2+2}}\right] + 9\bigg\} = \boxed{\dfrac{17}{2}}$

$iv) \displaystyle \lim_{x \to 2-} ax+b = \dfrac{17}{2} \Rightarrow \boxed{\boxed{2a + b = \dfrac{17}{2}}}$

$\textsf{--------------------}$$\textsf{--------------------}$$\textsf{--------------------}$$\textsf{--------------------}$

$v) \displaystyle \lim_{x \to 3-}\bigg\{(x-3)\cdot \left[\dfrac{\sqrt{x} - \sqrt{2} + \sqrt{x-2}}{\sqrt{x^2-4}}\right] + 9\bigg\} = \boxed{9}$

$\displaystyle vi) \lim_{x \to 3+} \dfrac{\sqrt[3]{3}\cdot(x-3)}{\sqrt[3]{x} - \sqrt[3]{3}} \cdot \dfrac{\sqrt[3]{3^2}+\sqrt[3]{3x} + \sqrt[3]{x^2}}{\sqrt[3]{3^2}+\sqrt[3]{3x} + \sqrt[3]{x^2}} \Longleftrightarrow \\\\\\\lim_{x \to 3+} \left[\sqrt[3]{3}\cdot(\sqrt[3]{3^2}+\sqrt[3]{3x} + \sqrt[3]{x^2})\right] = \boxed{9}$

Logo, resolvendo o sistema das equações de ii) e iv) e notando que v) e vi) resultam em c:

• a = 11/2
• b = -5/2
• c = 9