Limit of a fraction involving square and third roots.

Question

$$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$

Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multiplicate a polynomial by square and third roots?

Answer 1

Instead, write $\sqrt{n^2+1}= |n|\sqrt{1+\frac{1}{n^2}}$, and similar for other expression in the numerator. For the denominator, multiply both numerator and denominator. Then in the denominator you will have $n^2$, and in the numerator the term $$ (n^3+1)^{\frac{2}{3}} + ((n^3+1)(n^3+n^2+1))^{\frac{2}{3}} + (n^3+n^2+1)^{\frac{2}{3}} $$ Then, you can perform the same operation on them, e.g. on the first one it is $|n|(1+\frac{1}{n^3})^{\frac{1}{3}}$, and a lot of terms will cancel out

Answer 2

Yes, you can make that approach work. It's a little tedious and messy looking, but here's what you get:

$${\sqrt{n^2+1}-\sqrt{n^2+n}\over\sqrt[3]{n^3+1}-\sqrt[3]{n^3+n^2+1}}\\={(n^2+1)-(n^2+n)\over(n^3+1)-(n^3+n^2+1)}\cdot{(\sqrt[3]{n^3+1})^2+\sqrt[3]{n^3+1}\sqrt[3]{n^3+n^2+1}+(\sqrt[3]{n^3+n^2+1})^2\over\sqrt{n^2+1}+\sqrt{n^2+n}}\\ ={n-1\over n^2}\cdot{n^2\left(\left(\sqrt[3]{1+{1\over n^3}} \right)^2+\sqrt[3]{1+{1\over n^3}}\sqrt[3]{1+{1\over n}+{1\over n^3}}+\left(\sqrt[3]{1+{1\over n}+{1\over n^3}} \right)^2\right)\over n\left(\sqrt{1+{1\over n^2}}+\sqrt{1+{1\over n}} \right)}\\ ={n-1\over n}\cdot{\left(\sqrt[3]{1+{1\over n^3}} \right)^2+\sqrt[3]{1+{1\over n^3}}\sqrt[3]{1+{1\over n}+{1\over n^3}}+\left(\sqrt[3]{1+{1\over n}+{1\over n^3}} \right)^2\over \sqrt{1+{1\over n^2}}+\sqrt{1+{1\over n}}}\\ \to1\cdot{1^2+1\cdot1+1^2\over1+1}={3\over2}$$

The key is to factor an $n$ out of all the square and cube roots after converting the ratio of differences to a ratio of sums, after which the messy stuff that's left inside each of the roots simply tends to $1$.

Remark: I'm prone to making minus sign mistakes, so I always try to doublecheck that the sign of my answer makes sense. In this case it does: the numerator $\sqrt{n^2+1}-\sqrt{n^2+n}$ is negative, since $n^2+n$ is bigger than $n^2+1$, but so is the denominator, since $n^3+n^2+1$ is bigger than $n^3+1$, hence their ratio is positive, so the limit is non-negative, which agrees with the sign of the answer I got.

Answer 3

Considering $$y= \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$ compose Taylor series and use binomial expansions $$\sqrt{n^2+1}=n+\frac{1}{2 n}+\cdots$$ $$\sqrt{n^2+n}=n+\frac{1}{2}-\frac{1}{8 n}+\cdots$$ $$\sqrt[3]{n^3+1}=n+\frac{1}{3 n^2}+\cdots$$ $$\sqrt[3]{n^3+n^2+1}=n+\frac{1}{3}-\frac{1}{9 n}+\cdots$$ $$y=\frac {-\frac{1}{2}+\frac{5}{8 n}+\cdots } {-\frac{1}{3}+\frac{1}{9 n}+\cdots }\to \frac 32$$