Question

Calcule os limites

$b) \lim_{x \to \ 0} \frac{1- \sqrt{x-1} }{x} c) \lim_{x \to \ 1} \frac{ \sqrt{x+3} - 2 }{x-1} d) \lim_{x \to \ 0} \frac{ \ \sqrt{1-2x- x^{2} } - 1 }{x}$

Answer 1

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$\large\begin{array}{l} \textsf{Calcular os limites:}\\\\\\ \textsf{b) }\mathsf{\underset{x\to 0}{\ell im}~\dfrac{1-\sqrt{x-1}}{x}}\\\\\\ \textsf{Considere a fun\c{c}\~ao}\\\\ \mathsf{f(x)=\dfrac{1-\sqrt{x-1}}{x}} \end{array}$


$\large\textsf{Vamos encontrar o dom\'inio desta fun\c{c}\~ao. Devemos ter:}$

•   $\large\begin{array}{l} \mathsf{x-1\ge 0} \end{array}$

$\large\begin{array}{l} \mathsf{x\ge 1} \end{array}$


$\large\begin{array}{l} \textsf{O dom\'inio de f \'e }\mathsf{Dom(f)=[1,\,+\infty).}\\\\ \textsf{Observe que }\mathsf{x=0}\textsf{ n\~ao \'e um ponto de acumula\c{c}\~ao do dom\'inio}\\\textsf{de f, porque este \'e um ponto isolado do intervalo }\mathsf{[1,\,+\infty).} \end{array}$


$\large\begin{array}{l} \textsf{Portanto, n\~ao \'e poss\'ivel calcular o limite}\\\\ \mathsf{\underset{x\to 0}{\ell im}~\dfrac{1-\sqrt{x-1}}{x}}\\\\ \textsf{(o limite n\~ao existe).} \end{array}$

________


$\large\begin{array}{l} \textsf{c) }\mathsf{\underset{x\to 1}{\ell im}~\dfrac{\sqrt{x+3}-2}{x-1}}\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{\sqrt{x+3}-2}{x-1}\cdot \dfrac{\sqrt{x+3}+2}{\sqrt{x+3}+2}}\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(\sqrt{x+3}-2)\cdot (\sqrt{x+3}+2)}{(x-1)\cdot (\sqrt{x+3}+2)}} \end{array}$

$\large\begin{array}{l} \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{(\sqrt{x+3})^2-2^2)} {(x-1)\cdot (\sqrt{x+3}+2)}}\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{x+3-4}{(x-1)\cdot (\sqrt{x+3}+2)}}\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{x-1}{(x-1)\cdot (\sqrt{x+3}+2)}}\\\\ \mathsf{=\underset{x\to 1}{\ell im}~\dfrac{1}{\sqrt{x+3}+2}} \end{array}$

$\large\begin{array}{l} \mathsf{=\dfrac{1}{\sqrt{1+3}+2}}\\\\ \mathsf{=\dfrac{1}{\sqrt{4}+2}}\\\\ \mathsf{=\dfrac{1}{2+2}}\\\\ \mathsf{=\dfrac{1}{4}}\qquad\quad\checkmark \end{array}$

__________


$\large\begin{array}{l} \textsf{d) }\mathsf{\underset{x\to 0}{\ell im}~\dfrac{\sqrt{1-2x-x^2}-1}{x}}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{\sqrt{1-2x-x^2}-1}{x}\cdot \dfrac{\sqrt{1-2x-x^2}+1}{\sqrt{1-2x-x^2}+1}}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{(\sqrt{1-2x-x^2}-1)\cdot (\sqrt{1-2x-x^2}+1)}{x\cdot (\sqrt{1-2x-x^2}+1)}} \end{array}$

$\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{(\sqrt{1-2x-x^2})^2-1^2}{x\cdot (\sqrt{1-2x-x^2}+1)}}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{\diagup \hspace{-7} 1-2x-x^2-\diagup \hspace{-7} 1}{x\cdot (\sqrt{1-2x-x^2}+1)}}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{-\diagup \hspace{-8} x\cdot (2+x)}{\diagup \hspace{-8}x\cdot (\sqrt{1-2x-x^2}+1)}}\\\\ \end{array}$

$\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{-2+x}{\sqrt{1-2x-x^2}+1}}\\\\ \mathsf{=\dfrac{-2+0}{\sqrt{1-2\cdot 0-0^2}+1}}\\\\ \mathsf{=\dfrac{-2}{\sqrt{1}+1}}\\\\ \mathsf{=\dfrac{-2}{1+1}} \end{array}$

$\large\begin{array}{l} \mathsf{=\dfrac{-2}{2}}\\\\ \mathsf{=-1}\qquad\quad\checkmark \end{array}$


$\large\textsf{Bons estudos! :-)}$


Tags:  limite função irracional ponto de acumulação multiplicação pelo conjugado calcular resolver cálculo diferencial