What is the value of $$\lim_{x\to\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}$$ I tried to factorise the denominator or the numerator but it takes me nowhere. My teacher said it equals to $\frac{1}{2}$. But I don't get it. Can anyone help me?
2026-04-06 06:31:35.1775457095
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The limit of a fraction
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Factor out and simplify the highest power of $x$ in the denominator and in the numerator: $$\lim_{x\to\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1} = \lim_{x\to\infty}\frac{x^2\sqrt{1-\frac{1}{x^4}}}{x^2\left(2+\frac{3}{x}-\frac{1}{x^2}\right)} = \lim_{x\to\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}$$
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$$\frac{\sqrt{x^4-1}}{2x^2+3x-1}= \frac{x^2\sqrt{1-1/x^4}}{x^2(2+3/x-1/x^2)}= \frac{\sqrt{1-1/x^4}}{2+3/x-1/x^2}$$
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$$\lim_{x\rightarrow+\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}=\lim_{x\rightarrow+\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}=\frac{1}{2}$$ Also $$\lim_{x\rightarrow-\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}=\lim_{x\rightarrow-\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}=\frac{1}{2}$$ Thus, indeed, the answer is $\frac{1}{2}$.
Just dividing the numerator and denominator by $x^2$ gives $\frac{\sqrt{1-\frac{1}{x^4}}}{2-\frac{3}{x}-\frac{1}{x^2}}$. Now taking limit $x \rightarrow \infty $ gives the answer $\frac{1}{2}$.