اگر $x \to 0$، میتوانیم جای $\cos x$ از $1 - \frac{{{x^2}}}{2}$ استفاده کنیم:
$\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2 + 3x} - \sqrt {2 - x} }}{{\sqrt {1 - \cos x} }} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2 + 3x} - \sqrt {2 - x} }}{{\sqrt {1 - (1 - \frac{{{x^2}}}{2})} }}$
$ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2 + 3x} - \sqrt {2 - x} }}{{\sqrt {\frac{{{x^2}}}{2}} }} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2 + 3x} - \sqrt {2 - x} }}{{\frac{{|x|}}{{\sqrt 2 }}}}$
$ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {4 + 6x} - \sqrt {4 - 2x} }}{{ - x}}$
عبارت به دست آمده را در مزدوج صورت، ضرب و تقسیم میکنیم:
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4 + 6x} - \sqrt {4 - 2x} }}{{ - x}} \times \frac{{\sqrt {4 + 6x} + \sqrt {4 - 2x} }}{{\sqrt {4 + 6x} + \sqrt {4 - 2x} }}$
$ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{(4 + 6x) - (4 - 2x)}}{{ - x(\sqrt {4 + 6x} + \sqrt {4 - 2x} )}}$
$ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{8x}}{{ - x(2 + 2)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{8x}}{{ - 4x}} = - 2$