How to compute $\lim\limits_{x\to \infty} (3x-\sqrt{9x^2 - 7x})$

Question

I'm having a hard time simplifying this.

As usual I multiplied by the conjugate:

$$ (3x-\sqrt{9x^2 - 7x}) \cdot \left( \frac{3x+\sqrt{9x^2 - 7x}}{3x+\sqrt{9x^2 - 7x}}\right)$$

$$\frac{7x}{3x+\sqrt{9x^2-7x}}$$

Now I tried to divide all terms by $x$ by it still gives $\frac{1}{\sqrt{ \infty}}$. What am I missing?

Answer 1

You've done everything right so far. So we have $$\lim_{x\to \infty}\frac{7x}{3x+\sqrt{9x^2-7x}}$$

Now we factor out $x$ from $\sqrt{9x^2 - 7x} = \sqrt{x^2(9 - \frac 7x)}$ to get $|x|\sqrt{9 - \frac 7x}$. Now, given that $x$ is becoming arbitrarily large (positive), we take $|x| = x$.

$$\lim_{x\to \infty}\frac{7x}{3x+x\sqrt{9-\frac 7x}}$$

Factoring out the $x$ from numerator and denominator and canceling gives us:

$$\lim_{x\to \infty}\frac{7}{3+\sqrt{9-\frac 7x}} = \frac 76$$

Answer 2

$$ (3x-\sqrt{9x^2 - 7}) * \frac{(3x+\sqrt{9x^2 - 7})}{(3x+\sqrt{9x^2 - 7})}=\frac{7}{3x+\sqrt{9x^2-7}}\ne\frac{7x}{3x+\sqrt{9x^2-7}}$$