Proving $\lim_{n \to \infty}\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx=0$

Question

Prove the following: $$\lim_{n \to \infty}\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx=0$$

I tried to solve it using the Riemann integral definition with the partition $1/n$: $$\sum_{k=1}^{\infty}f(a+\frac{b-a}{n}k)\frac{b-a}{n}$$

But I got stuck. any suggestions or hints for how to approach it?

Answer 1

We have that

$$0\le \int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx \le \int_n^{n+1}\frac1{\sqrt{x}}dx=2(\sqrt{n+1}-\sqrt n)=\frac2{\sqrt{n+1}+\sqrt n}\to 0 $$

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As noticed in the comments, more simply

$$0\le \int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx \le \int_n^{n+1}\frac1{\sqrt{x}}dx\le \int_n^{n+1}\frac1{\sqrt{n}}dx=\frac1{\sqrt{n}}\to 0$$

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Another way, since $f(x)=\frac{\sqrt{x-1}}{\sqrt{x^2+1}} \to 0$ as $x\to \infty$ by definition $\forall \varepsilon>0\;\; \exists \delta\;\; \forall x\ge \delta \;\;f(x)\le\varepsilon$ then $\forall n\ge \delta$

$$\int_n^{n+1}f(x)\;dx \le \int_n^{n+1}\varepsilon \;dx =\varepsilon$$

and therefore by definition $f(x)=\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx\to 0$ which holds in general by the same argument for any $f(x)\to 0$.

Answer 2

Define $$F(x) := \int_1^x \frac{\sqrt{x-1}}{\sqrt{x^2+1}}\, dx$$ By the Fundamental Theorem of Calculus, $F$ is differentiable, and $$F'(x) = \frac{\sqrt{x-1}}{\sqrt{x^2+1}}$$ for all $x > 1$.

Choose any $n\in \Bbb N$, $n\ge 2$. Then $$\int_{n}^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}\, dx = F(n+1) - F(n) = \frac{F(n+1)-F(n)}{(n+1)-n}$$ By the Mean Value Theorem applied to $F$, there exists $c_n \in (n,n+1)$ such that $$ \frac{F(n+1)-F(n)}{(n+1)-n} = F'(c_n) = \frac{\sqrt{c_n-1}}{\sqrt{c_n^2+1}}$$ Using the bounds $n < c_n < n+1$, we have $$\frac{\sqrt{n-1}}{\sqrt{(n+1)^2+1}} < \frac{\sqrt{c_n-1}}{\sqrt{c_n^2+1}} < \frac{\sqrt{n}}{\sqrt{n^2+1}}$$ To summarize, $$\frac{\sqrt{n-1}}{\sqrt{(n+1)^2+1}} < \int_{n}^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}\, dx < \frac{\sqrt{n}}{\sqrt{n^2+1}}$$ Using the Squeeze Theorem with $$\lim_{n\to\infty} \frac{\sqrt{n-1}}{\sqrt{(n+1)^2+1}} = 0 \quad \text{and}\quad \lim_{n\to\infty} \frac{\sqrt{n}}{\sqrt{n^2+1}} = 0$$ it follows immediately that $$\lim_{n\to\infty} \int_{n}^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}} = 0$$

Answer 3

By the MVT for integrals, there is $c\in(n,n+1)$ such that $$\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx=\frac{\sqrt{c-1}}{\sqrt{c^2+1}}. $$ Note that if $n\to\infty$, $c\infty$ and $$ \lim_{c\infty}\frac{\sqrt{c-1}}{\sqrt{c^2+1}}=0$$ and hence $$ \lim_{n\infty}\int_n^{n+1}\frac{\sqrt{x-1}}{\sqrt{x^2+1}}dx=0. $$