calculates (without using L'Hopital) the following limit

Question

I don't understand how to calculate the limit $$\lim_{x\to7}\frac{1}{x-7}\int_7^x\sqrt{t^2+9}dt $$ without using the L'Hopital rule the picture.

Answer 1

Using $f(t)=\sqrt{t^2+9}$ $$\int_7^x f(t)dt=F(x)-F(7)$$ Then what you are asked to compute is $$\lim_{x\to 7}\frac{F(x)-F(7)}{x-7}=F'(x)|_{x=7}=f(7)$$

Answer 2

Hint: use mean value theorem. There exists $\xi \in [7,x]$ such that $$\int_7^x \sqrt{t^2+9}dt = (x-7) \sqrt{\xi^2+9}$$ You don't really need to know the value of $\xi$, the fact that $\xi \in [7,x]$ is enough.