Trigonometrik Fonksiyonlarda Limit

Trigonometrik fonksiyonlarda limit nasıl hesaplanır? Trigonometrik fonksiyonlarda limitin özellikleri, örnekler.

1-
$\displaystyle \underset{k\to 0}{\mathop{\lim }}\,\frac{\sin k}{k}=1$

2-
$\displaystyle \underset{k\to 0}{\mathop{\lim }}\,\frac{\tan k}{k}=1$

3-
$\displaystyle \underset{k\to 0}{\mathop{\lim }}\,\frac{\sin ak}{bk}=\frac{a}{b}$

4-
$\displaystyle \underset{k\to 0}{\mathop{\lim }}\,\frac{\tan ak}{bk}=\frac{a}{b}$

5-
$\displaystyle \underset{k\to 0}{\mathop{\lim }}\,\frac{\sin ak}{\sin bk}=\frac{a}{b}$

6-
$\displaystyle \underset{k\to 0}{\mathop{\lim }}\,\frac{\tan ak}{\tan bk}=\frac{a}{b}$

ÖRNEK:

$\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \frac{2x}{2}}{3{{x}^{2}}}$ limitini bulunuz.

$\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \frac{2x}{2}}{3{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( \frac{\sin \frac{x}{2}}{x/2} \right)}^{2}}.\frac{{{x}^{2}}}{4}}{3{{x}^{2}}}$

$\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \frac{2x}{2}}{3{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\left[ {{\left( \frac{\sin \frac{x}{2}}{x/2} \right)}^{2}}.\frac{{{x}^{2}}}{4}.\frac{1}{3{{x}^{2}}} \right]={{1}^{2}}.\frac{1}{12}=\frac{1}{12}$

ÖRNEK:
$\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1-\cos 2x \right)\tan x}{{{x}^{2}}\sin x}$ limitini bulalım.

$\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1-\cos 2x \right)\tan x}{{{x}^{2}}\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left[ 1-\left( 1-2{{\sin }^{2}}x \right) \right]\tan x}{{{x}^{2}}\sin x}$

$\displaystyle =\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x.\tan x}{{{x}^{2}}\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x.\tan x}{{{x}^{2}}}$

$\displaystyle =\underset{x\to 0}{\mathop{\lim }}\,\left( 2.\frac{\sin x}{x}\frac{\tan x}{x} \right)=2.1.1=2$ olur.

ÖRNEK:

$\displaystyle \underset{x\to a}{\mathop{\lim }}\,\frac{\sin x-\sin a}{{{x}^{2}}-{{a}^{2}}}$ limiti nedir?

$\displaystyle \underset{x\to a}{\mathop{\lim }}\,\frac{\sin x-\sin a}{{{x}^{2}}-{{a}^{2}}}=\underset{x\to a}{\mathop{\lim }}\,\frac{2.\cos \frac{x+a}{2}.\sin \frac{x-a}{2}}{{{x}^{2}}-{{a}^{2}}}$

$\displaystyle \underset{x\to a}{\mathop{\lim }}\,\left[ \frac{2\cos \left( \frac{x+a}{2} \right)}{x+a}.\frac{\sin \frac{x-a}{2}}{\left( \frac{x-a}{2} \right).2} \right]=\frac{2\cos a}{2a}.1.\frac{1}{2}=\frac{\cos a}{2a}$