Types of Infinity — Question 5

Question 5

Prove that $$\lim_{x\to\infty}\left(\sqrt{x^2+5x}-\sqrt{x^2+x}\right)=2.$$

Question 5 - Solution

We simplify the expression by factoring out $x$ from each square root.

Consider $$\sqrt{x^2+5x}-\sqrt{x^2+x}.$$

Factor $x^2$ inside each square root: $$\sqrt{x^2\left(1+\frac{5}{x}\right)}-\sqrt{x^2\left(1+\frac{1}{x}\right)}.$$

Since $x>0$ for large $x$, this becomes $$x\left(\sqrt{1+\frac{5}{x}}-\sqrt{1+\frac{1}{x}}\right).$$

Multiply and divide by the conjugate: $$x\cdot \frac{\left(\sqrt{1+\frac{5}{x}}-\sqrt{1+\frac{1}{x}}\right) \left(\sqrt{1+\frac{5}{x}}+\sqrt{1+\frac{1}{x}}\right)} {\sqrt{1+\frac{5}{x}}+\sqrt{1+\frac{1}{x}}}.$$

Simplify the numerator: $$\left(1+\frac{5}{x}\right)-\left(1+\frac{1}{x}\right) = \frac{4}{x}.$$

Thus the expression becomes $$x\cdot \frac{\frac{4}{x}}{\sqrt{1+\frac{5}{x}}+\sqrt{1+\frac{1}{x}}} = \frac{4}{\sqrt{1+\frac{5}{x}}+\sqrt{1+\frac{1}{x}}}.$$

Now take the limit as $x\to\infty$: $$\lim_{x\to\infty} \frac{4}{\sqrt{1+\frac{5}{x}}+\sqrt{1+\frac{1}{x}}} = \frac{4}{1+1} = 2.$$

$$\boxed{2}$$