Types of Infinity — Question 7

Question 7 - Solution

Begin with the expression $$x\bigl(\sqrt{x^2+1}-x\bigr).$$

Factor the expression inside the parentheses using a conjugate: $$\sqrt{x^2+1}-x = \frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x} = \frac{(x^2+1)-x^2}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}+x}.$$

Substitute this back: $$x\bigl(\sqrt{x^2+1}-x\bigr) = \frac{x}{\sqrt{x^2+1}+x}.$$

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

Thus, $$\frac{x}{x\sqrt{1+\frac{1}{x^2}}+x} = \frac{1}{\sqrt{1+\frac{1}{x^2}}+1}.$$

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

$$\boxed{\frac{1}{2}}$$