Types of Infinity — Question 4

Question 4

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

Question 4 - Solution

We simplify the expression using algebraic manipulation.

Consider $$x-\sqrt{x^2-3x}.$$

Multiply and divide by the conjugate: $$x-\sqrt{x^2-3x} = \frac{(x-\sqrt{x^2-3x})(x+\sqrt{x^2-3x})}{x+\sqrt{x^2-3x}}.$$

Simplify the numerator: $$x^2-(x^2-3x)=3x.$$

Thus, $$x-\sqrt{x^2-3x} = \frac{3x}{x+\sqrt{x^2-3x}}.$$

Factor $x$ out of the square root in the denominator: $$\sqrt{x^2-3x} = x\sqrt{1-\frac{3}{x}}.$$

Substitute into the expression: $$\frac{3x}{x+x\sqrt{1-\frac{3}{x}}} = \frac{3}{1+\sqrt{1-\frac{3}{x}}}.$$

Now take the limit as $x\to\infty$: $$\lim_{x\to\infty}\frac{3}{1+\sqrt{1-\frac{3}{x}}} = \frac{3}{1+\sqrt{1}} = \frac{3}{2}.$$

$$\boxed{\frac{3}{2}}$$