Solve this limit with elementary calculus Is there a way to compute this limit using only elementary properties of definite integrals and without using any advanced integration technique 
$$\lim_{h \to 0} \int_0^h \frac {\sqrt {9 + t^2}}{h}dt $$
Thanks in advance
 A: There is no need for trigonometric substitution, in fact you do not need a specific formula for an antiderivative at all.  Let $F$ be an antiderivative of the integrand in $g(t)=\sqrt{9+t^2}$.  Use the Mean Value Theorem to render 
$$\frac{F(h)-F(0)}{h}=g(u)$$
for $u$ within some range, which the theorem as written in your book will tell you.  As $h$ approaches zero the range gets pinched off and there can be only one possible value for the limit, no matter what specific form $F$ might take.
A: Take $$F(x) = \int_0^x \sqrt{9+t^2} dt$$ then we have by the definition of the derivative: $$F'(0) = \lim_{h\to 0} \frac{F(h) - F(0)}{h}$$
But $$\frac{F(h) - F(0)}{h} =  \int_0^h \frac {\sqrt {9 + t^2}}{h}dt$$
So we have: $$\lim_{h \to 0} \int_0^h \frac {\sqrt {9 + t^2}}{h}dt = F'(0)$$
But we know $$F'(x) = \sqrt{9 + x^2}$$ by the Fundamental theorem of calculus.
So $F'(0) = \ldots$
A: From
$$3\leq\sqrt{9+t^2}\leq\sqrt{9+t^2+{t^4\over36}}=3+{t^2\over6}\leq3+{h^2\over6}\qquad(0\leq t\leq h)$$
it follows that
$$3h\leq\int_0^h\sqrt{9+t^2}\>dt\leq \left(3+{h^2\over6}\right)\,h$$
and therefore
$$3\leq\int_0^h{\sqrt{9+t^2}\over h}\>dt\leq 3+{h^2\over6}\qquad(h>0)\ .$$
This shows that the limit in question is $3$.
