Question

$\lim_{x \to \ 7 } \frac{5- \sqrt{4+3x} }{7-x}$

Answer 1

$\underset{x \to 7}{\mathrm{\ell im}}\;\dfrac{5-\sqrt{4+3x}}{7-x}$


Faremos a seguinte mudança de variável:

$u=\sqrt{4+3x}\\ \\u^{2}=4+3x\\ \\ 3x=u^{2}-4\\ \\ x=\dfrac{u^{2}-4}{3}\\ \\ \\ x \to 7\;\;\Leftrightarrow\;\;u\to \sqrt{4+3\cdot 7}\\ \\ u \to \sqrt{25}\\ \\ u \to 5$


Trocando as variáveis no limite, temos

$\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{5-u}{7-\left(\frac{u^{2}-4}{3} \right )}\\ \\ =\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{3\cdot \left(5-u \right )}{3\cdot \left[7-\left(\frac{u^{2}-4}{3} \right ) \right ]}\\ \\ =\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{3\cdot \left(5-u \right )}{21-\left(u^{2}-4 \right )}\\ \\ =\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{3\cdot \left(5-u \right )}{21-u^{2}+4}\\ \\ =\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{3\cdot \left(5-u \right )}{25-u^{2}}\\ \\ =\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{3\cdot \left(5-u \right )}{\left(5+u \right )\cdot \left(5-u \right )}\\ \\ =\underset{u \to 5}{\mathrm{\ell im}}\;\dfrac{3}{5+u}\\ \\ =\dfrac{3}{5+\left(5 \right )}\\ \\ =\dfrac{3}{10}\\ \\ \\ \boxed{\underset{x \to 7}{\mathrm{\ell im}}\;\dfrac{5-\sqrt{4+3x}}{7-x}=\dfrac{3}{10}}$