Collection of Maths Problems

Differences of roots

Task number: 2868

• Variant 1

  \(\displaystyle \lim_{n\to\infty} \sqrt{n+5}-\sqrt{n-1}\)

• Hint

  Expand an expression of the form \(A - B\) into a fraction \(\frac{A+B}{A+B}\).

• Resolution

  \(\displaystyle \lim_{n\to\infty} \sqrt{n+5}-\sqrt{n-1} = \lim_{n\to\infty} (\sqrt{n+5}-\sqrt{n-1}) \frac{\sqrt{n+5} + \sqrt{n-1}}{\sqrt{n+5} + \sqrt{n-1}} =\)

  \[ =\lim_{n\to\infty} \frac{n+5 - (n-1)}{\sqrt{n+5} + \sqrt{n-1}} = \lim_{n\to\infty}\frac{6}{\sqrt{n+5} + \sqrt{n-1}} = 0 \]

  The last equality follows from the fact that the expression in the denominator goes to infinity.

• Result

  The limit equals \(0\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty} \sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2}\)

• Hint

  Expand an expression of the form \(A-B\) into a fraction \(\frac{A^2 + AB + B^2}{A^2 + AB + B^2}\).

• Resolution

  \(\displaystyle \lim_{n\to\infty} \sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2} =\\= \lim\limits_{n\to\infty} (\sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2}) \frac{(\sqrt[3]{(n+1)^2})^2 + \sqrt[3]{(n+1)^2}\sqrt[3]{(n-1)^2} + (\sqrt[3]{(n-1)^2})^2}{(\sqrt[3]{(n+1)^2})^2 + \sqrt[3]{(n+1)^2}\sqrt[3]{(n-1)^2} + (\sqrt[3]{(n-1)^2})^2} =\\= \lim\limits_{n\to\infty} \frac{(n+1)^2 - (n-1)^2}{\sqrt[3]{n^4 + 4n^3 + 6n^2 + 4n + 1} + \sqrt[3]{n^4 - 2n^2 + 1} + \sqrt[3]{n^4 - 4n^3 + 6n^2 - 4n + 1}} =\\= \lim\limits_{n\to\infty} \frac{4n}{n(\sqrt[3]{n + 4 + 6n^{-1} + 4n^{-2} + n^{-3}} + \sqrt[3]{n - 2n^{-1} + n^{-3}} + \sqrt[3]{n - 4 + 6n^{-1} - 4n^{-2}+ n^{-3}})} =\\= \lim\limits_{n\to\infty} \frac{4}{\sqrt[3]{n + 4 + 6n^{-1} + 4n^{-2} + n^{-3}} + \sqrt[3]{n - 2n^{-1} + n^{-3}} + \sqrt[3]{n - 4 + 6n^{-1} - 4n^{-2}+ n^{-3}}} = 0 \)

  For the last equality note that the expression in the denominator goes to infinity.

• Result

  The limit equals \(0\).

• Variant 3

  \(\displaystyle \lim_{n\to\infty} \sqrt{n}\left(\sqrt{n+1}-\sqrt{n} \right)\)

• Hint

  Expand an expression of the form \(A - B\) into a fraction \(\frac{A+B}{A+B}\).

• Resolution

  The solution begins like in variant 1, then we factor out \(\sqrt n\).

• Result

  The limit equals \(\displaystyle\frac 12\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty} \sqrt{n^2+n}-\sqrt{n}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \sqrt{n^2+n}-\sqrt{n}= \lim_{n\to\infty} \sqrt{n} (\sqrt{n-1}-1) \ge \lim_{n\to\infty} \sqrt{n} = \infty \)

  We have used the inequality \(\sqrt{n-1}-1\ge 1\) which holds for all \(n\ge 5\).

• Result

  The limit is \(\infty\).

• Variant 5

  \(\displaystyle \lim_{n\to\infty} \left(n\left(\sqrt{n-3}-n\right)\right)\)

• Resolution

  The root is defined for all \(n\ge 3\).

  \(\displaystyle \lim_{n\to\infty} \left(n\left(\sqrt{n-3}-n\right)\right)= \lim_{n\to\infty} \left(n^2\left(\sqrt{\frac{1}{n}-\frac{3}{n^2}}-1\right)\right)= \infty\left(\sqrt{0-0}-1\right)=-\infty \)

• Result

  The limit equals \(-\infty\).

• Variant 6

  \(\displaystyle \lim_{n\to\infty} \left(\sin\sqrt{n+1}- \sin\sqrt{n}\right)\)

• Hint

  Use a sum-to-product identity and estimate the sine function.

• Resolution

  \(\displaystyle \lim_{n\to\infty} \left(\sin\sqrt{n+1}- \sin\sqrt{n}\right)\le \lim_{n\to\infty} \left|\sin\sqrt{n+1}- \sin\sqrt{n}\right|=\)

  \[ =\lim_{n\to\infty} \left|2\cos\left(\frac{\sqrt{n+1}+\sqrt{n}}2\right) \sin\left(\frac{\sqrt{n+1}-\sqrt{n}}2\right)\right|\le\] \[ \le\lim_{n\to\infty} 2 \sin\left(\frac{\sqrt{n+1}-\sqrt{n}}2\right)\le \lim_{n\to\infty} \left(\sqrt{n+1}-\sqrt{n}\right) = 0 \]

  The last equality follows from an argument like in variant 1.

• Result

  The limit equals \(0\).