Правило на Лопитал

[Petia Jordanova]

1) lim e^x.sinx+e^x.cosx-1 / 6x+5x^4
x->o


2) lim e^-x.sinx-x / 3x^2-x^5
x->o

3) lim √1+x^2 (cqloto e v √) -1 / x^2
x->o

4) lim x. [e^2/x-1]
x->+∞

[kmitov]

[tex]\lim_{x \to 0}\frac{ e^x.\sin x+e^x.\cos x-1}{6x+5x^4}=\left[\frac{0}{0}\right]=l'Hospital=[/tex]
[tex]\lim_{x \to 0}\frac{ e^x.\sin x+e^x.\cos x+e^x \cos x-e^x \sin x}{6+20x^3}=\frac{2}{6}=\frac{1}{3}.[/tex]



[tex]\lim_{x \to 0}\frac{e^{-x}.\sin x-x}{3x^2-x^5}=\left[\frac{0}{0}\right]=l'Hospital=\lim_{x \to 0}\frac{-e^{-x}.\sin x+e^{-x}\cos x-1}{6x-5x^4}=\left[\frac{0}{0}\right]=l'Hospital=[/tex]
[tex]\lim_{x \to 0}\frac{e^{-x}.\sin x-e^{-x}\cos x-e^{-x}\cos x+e^{-x}(-\sin x)}{6-20x^3}=-\frac{1}{3}[/tex]


[tex]\lim_{x \to 0}\frac{\sqrt{1+x^2}-1}{ x^2}=\left[\frac{0}{0}\right]=l'Hospital=[/tex]
[tex]\lim_{x \to 0}\frac{\frac{1}{2\sqrt{1+x^2}}2x}{2x}=\lim_{x \to 0}\frac{1}{2\sqrt{1+x^2}=\frac{1}{2}.[/tex]



[tex]\lim_{x \to \infty} x. [e^{2/x}-1]=[\infty.0]=\lim_{x \to \infty} \frac{e^{2/x}-1}{1/x}=\left[\frac{0}{0}\right]=l'Hospital=[/tex]
[tex]\lim_{x \to \infty} \frac{e^{2/x}(-2/x^2)}{-1/x^2}=\lim_{x \to \infty} 2e^{2/x}=2e^0=2[/tex]