Question

4 si 5 va rog frumos !

Answer 1

4)

Merge rezolvat foarte usor cu regula lui L'Hopital :

$\lim_{x \to 1} \frac{\sqrt{x+3}+\sqrt{2x-1}-3}{x^{2}-1}=\lim_{x \to 1} \frac{\frac{1}{2\sqrt{x+3}}+\frac{1}{\sqrt{2x-1}}}{2x} = \frac{1+\frac{1}{4}}{2} = \frac{5}{8}$

5)

Observam ca ne aflam in cazul de nederminare $0^{0}$

Vom regula exponentului : $a^{x} = e^{ln(a^{x})} = e^{xln(a)}$

Asadar limita devine:

$\lim_{x \to \frac{\pi}{2}} e^{\frac{1}{2x-\pi}ln(sin(x))}$

In acest moment ne intereseaza sa aflam limita exponentului.

$\lim_{x \to \frac{\pi}{2}} \frac{1}{2x-\pi}ln(sin(x)) = \lim_{x \to \frac{\pi}{2}} \frac{ln(sin(x))}{2x-\pi}$

Aplicand L'Hopital obtinem:

$\lim_{x \to \frac{\pi}{2}} \frac{ctg(x)}{2} = \frac{ctg(\frac{\pi}{2})}{2} = 0$

In concluzie, limita noastra este:

$e^{0} = 1$