ANPEC 2003 - Q7

por temujin Qui Abr 11, 2013 11:41 am

(0) $\small \lim_{x \to 5}\frac{4x^2-100}{x-5}=40$

V. Por L'Hospital:

$\small \lim_{x \to 5}\frac{4x^2-100}{x-5}=\lim_{x \to 5} \frac{8x}{1}=40$

(1) $\small \lim_{x \to \infty}\left ( x^2e^{1/x} -x^2\right )=0$

F. Fatorando o polinômio:

$\small \lim_{x \to \infty}x^2\left ( e^{1/x}-1 \right )=\lim_{x \to \infty} \frac{e^{1/x}-1}{\frac{1}{x^2}}$

E aplicando L'Hospital:

$\small \lim_{x \to \infty} \frac{e^{1/x}-1}{\frac{1}{x^2}}= \frac{e^{1/x}}{\frac{-2}{x^3}}=\infty$

(2) $\small \lim_{x \to 1}\frac{x ln(x)}{x^2-1}=1$

F.

$\small \lim_{x \to 1}\frac{x ln(x)}{x^2-1}=\lim_{x \to 1}\frac{ln(x)+1}{2x}=\frac{1}{2}$

(3) Se $\small cos(x+A)=\frac{\sqrt{3}}{2}cos x-\frac{1}{2}sen x, \ entao \ A=-\frac{\pi}{6}$ .

F.

$\small \\ cos(x+A)=cos(x)cos)(A)-sen(x)sen(A)=cos(x)\frac{\sqrt{3}}{2}-sen(x)\frac{1}{2}\\$

Logo, $\small cos(A)=\frac{\sqrt{3}}{2} \ ; \ sen(A)=\frac{1}{2} \Rightarrow A=\frac{\pi}{6}$

(4) $\small \lim_{n \to \infty} \frac{-1+\left (cos\left ( \frac{\pi x}{2} \right ) \right )^n}{1+\left (cos\left ( \frac{\pi x}{2} \right ) \right )^n}=-1, 0 < x < 1$

V.

$\small 0 < cos \left ( \frac{\pi x}{2} \right ) < 1 \Rightarrow \lim_{n \to \infty} \frac{-1+\left (cos\left ( \frac{\pi x}{2} \right ) \right )^n}{1+\left (cos\left ( \frac{\pi x}{2} \right ) \right )^n}=-1,$