[tex] \lim_{x \to \infty} \frac{ \sqrt{2 x^{2} +2x-3}- \sqrt{2 x^{2} -2x-2} }{2} = ...?[/tex]

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[tex] \lim_{x \to \infty} \frac{ \sqrt{2 x^{2} +2x-3}- \sqrt{2 x^{2} -2x-2} }{2} [/tex]

[tex]= \lim_{x \to \infty} \frac{ \sqrt{2 x^{2} +2x-3}- \sqrt{2 x^{2} -2x-2} }{2} \times \frac{\sqrt{2 x^{2} +2x-3}+ \sqrt{2 x^{2} -2x-2}}{\sqrt{2 x^{2} +2x-3}+ \sqrt{2 x^{2} -2x-2}} [/tex]

[tex]= \lim_{x \to \infty} \frac{ 2 x^{2} +2x-3- (2 x^{2} -2x-2)}{2(\sqrt{2 x^{2} +2x-3}+ \sqrt{2 x^{2} -2x-2})} [/tex]

[tex]= \lim_{x \to \infty} \frac{ 2 x^{2} +2x-3- 2 x^{2} +2x+2}{2(\sqrt{2 x^{2} +2x-3}+ \sqrt{2 x^{2} -2x-2})} [/tex]

[tex]= \lim_{x \to \infty} \frac{4x-1}{2(\sqrt{2 x^{2} +2x-3}+ \sqrt{2 x^{2} -2x-2})} [/tex]

[tex]= \lim_{x \to \infty} \frac{4x}{2(\sqrt{2 x^{2}}+ \sqrt{2 x^{2}})} \,\,\,\,\,\,\,[diambil\,\,suku\,\,dengan\,\,pangkat\,\,tertinggi][/tex]

[tex]= \lim_{x \to \infty} \frac{4x}{2(x\sqrt{2}+ x\sqrt{2})} [/tex]

[tex]= \lim_{x \to \infty} \frac{4x}{2x\sqrt{2} \,\,+\,\, 2x\sqrt{2}} [/tex]

[tex]= \lim_{x \to \infty} \frac{4x}{4x\sqrt{2}} [/tex]

[tex]= \lim_{x \to \infty} \frac{4}{4\sqrt{2}} [/tex]

Sehingga :
[tex]= \frac{4}{4\sqrt{2}} [/tex]

[tex]= \frac{4}{4\sqrt{2}} \times \frac{4 \sqrt{2} }{4 \sqrt{2}} [/tex]

[tex]= \frac{16 \sqrt{2} }{16 \times 2} [/tex]

[tex]= \frac{16 \sqrt{2} }{32} [/tex]

[tex]= \frac{\sqrt{2} }{2} [/tex]