teman, tolong bantu saya soal ituu yaaa

[tex]\displaystyle \lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}\cdot\frac{\frac1x}{\frac1x}\\\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\lim_{x\to\infty}\frac{2-\frac1x}{\frac1x\sqrt{2x^2+3}}\\\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\lim_{x\to\infty}\frac{2-\frac1x}{\sqrt{2+\frac3{x^2}}}\\\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\frac{2-\frac1\infty}{\sqrt{2+\frac3{\infty^2}}}[/tex]
[tex]\displaystyle \lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\frac{2-0}{\sqrt{2+0}}\\\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\frac{2}{\sqrt{2}}\\\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt2}{\sqrt2}\\\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\frac{2\sqrt{2}}{2}\\\boxed{\boxed{\lim_{x\to\infty}\frac{2x-1}{\sqrt{2x^2+3}}=\sqrt{2}}}[/tex]