4)
Merge rezolvat foarte usor cu regula lui L'Hopital :
[tex]\lim_{x \to 1} \frac{\sqrt{x+3}+\sqrt{2x-1}-3}{x^{2}-1}=\lim_{x \to 1} \frac{\frac{1}{2\sqrt{x+3}}+\frac{1}{\sqrt{2x-1}}}{2x} = \frac{1+\frac{1}{4}}{2} = \frac{5}{8}[/tex]
5)
Observam ca ne aflam in cazul de nederminare [tex]0^{0}[/tex]
Vom regula exponentului : [tex]a^{x} = e^{ln(a^{x})} = e^{xln(a)}[/tex]
Asadar limita devine:
[tex]\lim_{x \to \frac{\pi}{2}} e^{\frac{1}{2x-\pi}ln(sin(x))}[/tex]
In acest moment ne intereseaza sa aflam limita exponentului.
[tex]\lim_{x \to \frac{\pi}{2}} \frac{1}{2x-\pi}ln(sin(x)) = \lim_{x \to \frac{\pi}{2}} \frac{ln(sin(x))}{2x-\pi}[/tex]
Aplicand L'Hopital obtinem:
[tex]\lim_{x \to \frac{\pi}{2}} \frac{ctg(x)}{2} = \frac{ctg(\frac{\pi}{2})}{2} = 0[/tex]
In concluzie, limita noastra este:
[tex]e^{0} = 1[/tex]
