Find the Limit of √(9x² − x) − 3x as x → ∞

The limit in the title is $\lim_{x\to\infty}\left(\sqrt{9x^2-x}-3x\right).$ Its value is $-1/6$. The key is to rationalize the difference rather than try to evaluate its two large terms separately.

Why direct substitution does not work

As $x$ grows, both $\sqrt{9x^2-x}$ and $3x$ grow approximately like $3x$. Their difference is therefore an indeterminate form of type $\infty-\infty$: knowing that each term grows without bound does not tell you what their difference approaches.

For sufficiently large positive $x$, compare the terms by squaring: $9x^2-x<9x^2$, so $\sqrt{9x^2-x}<3x$. The expression is negative, but this comparison alone does not determine its limit. Rationalization reveals the size of the difference.

Rationalize using the conjugate

The conjugate of $\sqrt{9x^2-x}-3x$ is $\sqrt{9x^2-x}+3x$. Multiply by this quantity in both the numerator and denominator. Their product uses the difference-of-squares identity, $(a-b)(a+b)=a^2-b^2$:

$$\begin{aligned} \sqrt{9x^2-x}-3x &=\frac{(\sqrt{9x^2-x}-3x)(\sqrt{9x^2-x}+3x)}{\sqrt{9x^2-x}+3x}\\ &=\frac{(9x^2-x)-9x^2}{\sqrt{9x^2-x}+3x}\\ &=\frac{-x}{\sqrt{9x^2-x}+3x}. \end{aligned}$$

Since the limit is taken as $x\to\infty$, we can divide the numerator and denominator by the positive quantity $x$:

$$\frac{-x}{\sqrt{9x^2-x}+3x} =\frac{-1}{\sqrt{9-1/x}+3}.$$

Now $1/x\to0$, so the denominator approaches $\sqrt{9}+3=6$. Therefore,

$$\lim_{x\to\infty}\left(\sqrt{9x^2-x}-3x\right)=-\frac{1}{6}.$$

Check with a concrete value

At $x=100$, the original expression is $\sqrt{89900}-300$, which is about $-0.1667$. The rationalized form gives

$$-\frac{1}{\sqrt{9-1/100}+3} \approx -0.1667.$$

This numerical check is consistent with the exact limit $-1/6\approx-0.1667$. A numerical check cannot prove a limit, but it can help catch a sign or algebra error.

Common mistakes

• Treating $\infty-\infty$ as zero. The terms become large at similar rates, but their difference can approach a nonzero number, diverge, or behave in other ways. Simplify the difference first.
• Dropping the minus sign. The numerator after applying the difference-of-squares identity is $(9x^2-x)-9x^2=-x$, not $x$.
• Using $\sqrt{x^2}=x$ without checking the sign. In general, $\sqrt{x^2}=|x|$. Here $x\to+\infty$, so $|x|=x$; that is why factoring out $x$ in the limit is valid.
• Assuming the same result for $x\to-\infty$. For negative $x$, $\sqrt{x^2}=|x|=-x$. In fact, $\sqrt{9x^2-x}-3x$ grows without bound as $x\to-\infty$, rather than approaching $-1/6$.

If you meant a fraction instead

If you instead meant $\lim_{x\to\infty}\frac{\sqrt{9x^2-x}}{3x}$, its limit is $1$, not $-1/6$. For $x>0$, divide the numerator and denominator by $x$ to get $\frac{\sqrt{9-1/x}}{3}\to1$. Parentheses and a fraction bar change the problem, so check the original formatting if this interpretation does not match what you intended.