Types of Infinity — Question 3

Question 3

Prove that $$\lim_{x\to\infty}\sqrt{x^2+x}-x=\frac{1}{2}.$$

Question 3 - Solution

We begin by algebraically simplifying the expression.

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

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

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

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

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

Substitute: $$\frac{x}{x\sqrt{1+\frac{1}{x}}+x} = \frac{x}{x\left(\sqrt{1+\frac{1}{x}}+1\right)}.$$

Cancel $x$: $$\frac{1}{\sqrt{1+\frac{1}{x}}+1}.$$

Now take the limit as $x\to\infty$: $$\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x}}+1} = \frac{1}{\sqrt{1+0}+1} = \frac{1}{2}.$$

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