Tìm các đường tiệm cận ngang của đồ thi hàm số y=(7sqrt (x^2+3)+4)/(x) y=7 và y=-7 y=-7 y=+infty y=7

To find the horizontal asymptotes of the function $y=\frac{7\sqrt{x^2+3}+4}{x}$, we need to evaluate the limits as $x$ approaches positive and negative infinity.<br /><br />Let's consider the limit as $x \to \infty$:<br />$$ \lim_{x \to \infty} \frac{7\sqrt{x^2+3}+4}{x} $$<br />We can rewrite the expression as:<br />$$ \lim_{x \to \infty} \frac{7|x|\sqrt{1+\frac{3}{x^2}}+4}{x} $$<br />Since $x \to \infty$, $x > 0$, so $|x| = x$. Then:<br />$$ \lim_{x \to \infty} \frac{7x\sqrt{1+\frac{3}{x^2}}+4}{x} = \lim_{x \to \infty} \left(7\sqrt{1+\frac{3}{x^2}} + \frac{4}{x}\right) = 7\sqrt{1+0} + 0 = 7 $$<br />Now let's consider the limit as $x \to -\infty$:<br />$$ \lim_{x \to -\infty} \frac{7\sqrt{x^2+3}+4}{x} $$<br />Since $x \to -\infty$, $x < 0$, so $|x| = -x$. Then:<br />$$ \lim_{x \to -\infty} \frac{7(-x)\sqrt{1+\frac{3}{x^2}}+4}{x} = \lim_{x \to -\infty} \left(-7\sqrt{1+\frac{3}{x^2}} + \frac{4}{x}\right) = -7\sqrt{1+0} + 0 = -7 $$<br />Therefore, the horizontal asymptotes are $y = 7$ and $y = -7$.<br /><br />Final Answer: The final answer is $\boxed{y=7 và y=-7}$