Use the Fundamental Theorem of Calculus I to evaluate $\lim_{x \to 0} \frac{1}{x} \int_0^x \sqrt{9+t^2}dt$

Question

Use the Fundamental Theorem of Calculus I to evaluate $\lim_{x \to 0} \dfrac{1}{x} {\displaystyle \int}_0^x \sqrt{9+t^2}dt$

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My professor sets $F(x) = \int_0^x \sqrt{9+t^2}dt$

We also find that,

$$\lim_{x \to 0} \frac{1}{x} \int_0^x \sqrt{9+t^2}dt = F'(0)$$

And thus,

$$\lim_{x \to 0} \frac{1}{x} \int_0^x \sqrt{9+t^2}dt = f(0) = \sqrt{9 + 0^2}$$

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But where does the $\frac{1}{x}$ go? It seems as if it was ignored. What if we had $x^4$ instead of just $x$? I hope this isn't something terribly trivial that I'm somehow missing. Thank you in advance.

Answer 1

If $F(x) = \int_0^x \sqrt{9+t^2} \, dt$, then $$F'(0) = \lim_{x \to 0} \frac{F(x)-F(0)}{x}=\lim_{x\to0}\frac{1}{x} \left(\int_0^x\sqrt{9+t^2}\,dt-0\right).$$