☰ matematika1

Neodređeni oblik FUNKCIJE REALNE VARIJABLE Primjena kada

Primjena $\lim (\sin x/x)$ kada $x\to 0$

Izračunajte:

a): $\displaystyle\lim_{x\to\pm\infty}x\sin{\frac{1}{x}}$ ,
b): $\displaystyle\lim_{x\to0}\frac{2\arcsin{x}}{3x}$ ,
c): $\displaystyle\lim_{x\to0}\frac{\mathop{\mathrm{tg}}\nolimits {x}-\sin{x}}{x^3}$ ,
d): $\displaystyle\lim_{x\to0}\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin^2 x}$ .

Rješenje. Ideja u ovim zadacima je transformirati izraz pod limesom tako da možemo primijeniti jednakost

$\displaystyle \lim_{x\to 0} \frac{\sin (ax)}{x}=a,$

(4.1)

koja se dobije iz

[M1, primjer 4.6] supstitucijom

a)

Supstitucijom $\displaystyle \frac{1}{x}=t$ dobivamo

$\displaystyle \lim_{x\to\pm\infty}x\sin{\frac{1}{x}}= \begin{Bmatrix} \displa... ...\to\pm\infty\\ t\to 0\pm \end{Bmatrix}= \lim_{t\to 0\pm} \frac{\sin t}{t}=1.$

b)

Supstitucijom $\arcsin x=t$ i primjenom

[M1, teorem 4.3] dobivamo

$\displaystyle \lim_{x\to0}\frac{2\arcsin{x}}{3x}= \begin{Bmatrix} \arcsin x=t... ...ac{2}{3}\cdot\frac{1}{\displaystyle\lim_{t\to 0} \frac{\sin t}{t}}=\frac{2}{3}.$

c)

Primjenom identiteta $\mathop{\mathrm{tg}}\nolimits x=\frac{\sin x}{\cos x}$ i $\sin^2 \frac{x}{2}=\frac{1}{2}(1-\cos x)$ dobivamo

$\displaystyle \lim_{x\to0}\frac{\mathop{\mathrm{tg}}\nolimits {x}-\sin{x}}{x^3}$	$\displaystyle = \lim_{x\to0}\frac{\sin x(1-\cos{x})}{x^3\cos x}=$
	$\displaystyle =\lim_{x\to0}\frac{2\sin x \sin^2\frac{x}{2}}{x^3\cos x}=$
	$\displaystyle =\lim_{x\to0}2\cdot\frac{\sin x}{x}\cdot\frac{\sin^2\frac{x}{2}}{x^2}\cdot\frac{1}{\cos x}.$

Zbog neprekidnosti kvadratne funkcije i

[M1, teorem 4.7 (ii)] vrijedi

$\displaystyle \lim_{x\to0}\frac{\sin^2\frac{x}{2}}{x^2}=\lim_{x\to0}{\left(\fra... ...}\frac{\sin \frac{x}{2}}{x}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}.$

Prema

[M1, teorem 4.3] sada slijedi

$\displaystyle \lim_{x\to0}\frac{\mathop{\mathrm{tg}}\nolimits {x}-\sin{x}}{x^3}... ...^2}\cdot\lim_{x\to0}\frac{1}{\cos x}=2\cdot1\cdot\frac{1}{4}\cdot1=\frac{1}{2}.$

d)

Racionalizacijom brojnika, primjenom identiteta $\sin^2 \frac{x}{2}=\frac{1}{2}(1-\cos x)$ te korištenjem

[M1, teorem 4.3] i

[M1, teorem 4.7 (ii)], dobivamo

	$\displaystyle \quad\,\lim_{x\to 0}\frac {\sqrt{2}-\sqrt{1+\cos x}}{\sin^2 x} = ... ...cos x}}{\sin^2 x}\cdot\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}$
	$\displaystyle = \lim_{x\to 0}\frac{2-(1+\cos x)}{\sin^2 x\cdot (\sqrt{2}+\sqrt{... ...os x})}= \lim_{x\to 0}\frac{1-\cos x}{\sin^2 x\cdot (\sqrt{2}+\sqrt{1+\cos x})}$
	$\displaystyle = \lim_{x\to 0} \frac{2\sin^2\left(\frac{x}{2}\right)}{\sin^2 x\c... ...\displaystyle\frac{\sin x}{x}\right)^2}\cdot \frac{1}{\sqrt{2}+\sqrt{1+\cos x}}$
	$\displaystyle = 2\cdot\frac{\displaystyle\left(\lim_{x\to 0}\frac{\sin\frac{x}{... ...\left(\frac{1}{2}\right)^2}{1^2}\cdot\frac{1}{2\sqrt{2}} =\frac{1}{4 \sqrt{2}}.$

Neodređeni oblik FUNKCIJE REALNE VARIJABLE Primjena kada