Types of Infinity — Question 6

Question 6

Prove that $$\lim_{x\to\infty}\left(\sqrt{x^2+4x+1}-x\right)=2.$$

Question 6 - Solution

Start with the expression $$\sqrt{x^2+4x+1}-x.$$

Multiply and divide by the conjugate: $$\sqrt{x^2+4x+1}-x = \frac{(\sqrt{x^2+4x+1}-x)(\sqrt{x^2+4x+1}+x)}{\sqrt{x^2+4x+1}+x}.$$

Simplify the numerator: $$(x^2+4x+1)-x^2=4x+1.$$

Thus, $$\sqrt{x^2+4x+1}-x = \frac{4x+1}{\sqrt{x^2+4x+1}+x}.$$

Factor $x$ from the square root in the denominator: $$\sqrt{x^2+4x+1} = x\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}.$$

Substitute: $$\frac{4x+1}{x\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+x} = \frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1}.$$

Take the limit as $x\to\infty$: $$\lim_{x\to\infty} \frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1} = \frac{4}{1+1} = 2.$$

$$\boxed{2}$$