Question

hallar los límites de cada una.​

Answer 1

Respuesta:

4. $2\sqrt{3}}$

5. $\frac{1}{4}$

6. $0$

7. $\frac{11}{2}$

Explicación paso a paso:

4. $\lim _{x\to 7}\left\frac{x-7}{\:\sqrt{x-4}-\sqrt{3}}\right \longrightarrow \left\frac{(x-7)(\sqrt{x-4}+\sqrt{3})}{(\:\sqrt{x-4}-\sqrt{3})(\sqrt{x-4}+\sqrt{3})}\right \longrightarrow \lim _{x\to 7}\sqrt{x-4}+\sqrt{3} \longrightarrow \\\\ \sqrt{7-4}+\sqrt{3} \longrightarrow \left\sqrt{3}+\sqrt{3}} \longrightarrow \left2\sqrt{3}$

5. $\lim _{x\to 4}\frac{\sqrt{x}-2}{\:x-4}\right \longrightarrow \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} \longrightarrow \lim _{x\to 4}\frac{1}{\sqrt{x} +2} \longrightarrow \frac{1}{\sqrt{4} +2} \longrightarrow \frac{1}{4}$

6. $\lim _{x\to -4}\left\frac{x+4}{\:\sqrt{x+4}}\right \longrightarrow \lim _{x\to -4}\left{\:\sqrt{x+4}}\right \longrightarrow \sqrt{-4+4} =0$

7. $\lim _{x\to \infty }\left(\sqrt{x^2+x+2}-\left(x-5\right)\right) \longrightarrow \lim _{x\to \infty \:}\left(\frac{11x-23}{\sqrt{x^2+x+2}+x-5}\right) \longrightarrow \\\\\frac{\lim _{x\to \infty \:}\left(11-\frac{23}{x}\right)}{\lim _{x\to \infty \:}\left(\sqrt{1+\frac{1}{x}+\frac{2}{x^2}}+1-\frac{5}{x}\right)} \longrightarrow \frac{11}{2}$