$\lim\limits_{h\to0}\frac{1}{h}\int_1^{1+h}\sqrt{1+t^2}dt$

Question

I need to find $\lim\limits_{h\to0}\frac{1}{h}\int_1^{1+h}\sqrt{1+t^2}dt$

I try to

$step1$

$\lim\limits_{h\to0}\frac{1}{h}[\frac{2}{3}(1+t^2)^\frac{3}{2}.\frac{1}{3}t^3]\mid_0^{1+h}$

$step2$

$\lim\limits_{h\to0}\frac{1}{h}[\frac{2}{3}(1+(1+h)^2)^\frac{3}{2}.\frac{1}{3}(1+h)^3]$

$step3$

$\lim\limits_{h\to0}\frac{1}{h}[\frac{2}{3}(1+(1+h)^3).\frac{1}{3}(1+h)^3]$

I not sure these steps is right?

Answer 1

Note that: $$\int\sqrt{1+t^2}dt=\frac12\left(t\sqrt{1+t^2}+\ln(t+\sqrt{1+t^2})\right).$$ Plugging $1+h$ and "1" and estimating the limit is quite complicated. Instead, follow the suggestion given in the comment.

Note that: $$\lim\limits_{h\to0}\frac{1}{h}\int_1^{1+h}\sqrt{1+t^2}dt=\lim\limits_{h\to0}\frac{\int_1^{1+h}\sqrt{1+t^2}dt}{h}\overbrace{=}^{LR}\lim\limits_{h\to0}\frac{\sqrt{1+(1+h)^2}}{1}=\sqrt{2}.$$

Answer 2

In general way, let $X$ be a Banach space and $f\colon [a,b] \longrightarrow X$ a continuous function. Then $$\lim_{h\to 0}\frac{1}{h}\int_c^{c+h} f(t)dt=f(c),$$ for all $c\in [a,b)$. It suffices to see that \begin{align*} \left\Vert\frac{1}{h}\int_c^{c+h} f(t)dt-f(c)\right\Vert &=& \left\Vert\frac{1}{h}\int_c^{c+h} (f(t)-f(c)) dt \right\Vert &\leq& \frac{1}{h}\int_c^{c+h} \|f(t)-f(c)\| dt &\leq& \sup\limits_{t\in [c,c+h]} \|f(t)-f(c)\| \to 0. \end{align*}

Answer 3

By the mean value theorem, there is $x(h) \in [1,1+h]$ such that

$$\int_1^{1+h}\sqrt{1+t^2}dt=h \sqrt{1+x(h)^2}.$$

Hence

$$\lim\limits_{h\to0}\frac{1}{h}\int_1^{1+h}\sqrt{1+t^2}dt=\sqrt{2}.$$

Answer 4

Correct me if wrong:

$\sqrt{1+t^2}$is continuos in $\mathbb{R}.$

Then:

$F(x) := \int_{1}^{x} \sqrt{1+t^2}dt$ is

differentiable,and we have

$F'(x)= \sqrt{1+x^2}.$

$F'(1)= \lim_{h \rightarrow 0} \dfrac {\int_{1}^{1+h} \sqrt{1+t^2}dt -0}{h} =$

$\sqrt{1+1} =√2.$