# Calculate $\lim_{x\to0}{F(x)\over g(x)}$, where $g(x)=x$ and $F(x)=\int_0^x {e^{2t}-2e^t+1\over 2\cos3t-2\cos2t+\cos t} \, dt$.

Calculate $\displaystyle\lim_{x\to0}{F(x)\over g(x)}$, where $g(x)=x$ and $\displaystyle F(x)=\int_0^x {e^{2t}-2e^t+1\over 2\cos3t-2\cos2t+\cos t} \, dt$.

i'd love for someone to explain not only the technical procedure here but also what are the theorems that allow it to take place. To my understanding, the Fundamental theorem of calculus is key here and Newton-Leibniz theorem also contributes. Even though I do feel I understand them as well as the subtle connection they made between antiderivatives and definite integrals I can't seem to apply any of that when it comes to limits and proofs.

Note that the problem is just a trivial application of Fundamental Theorem of Calculus. Let $f(t)$ be the integrand given in question and we have $F(x) = \int_{0}^{x}f(t)\,dt$. Note that $F(0) = 0$ and $g(x) = x$ so $$\lim_{x \to 0}\frac{F(x)}{g(x)} = \lim_{x \to 0}\frac{F(x) - F(0)}{x} = F'(0)$$ Since integrand $f(t)$ is continuous at $t = 0$ it follows by Fundamental Theorem of Calculus that $F'(0)$ exists and is equal to $f(0) = 0$. I don't see here any need for the use of L'Hospital's Rule.

BTW the large expression for the integrand $f(t)$ is only given here to intimidate the student and one should use the intimidation shield by just writing it symbolically as $f(t)$ (it saves typing effort too).

• Wow! I see! Thanks u very much:-) – Rubenz Jun 23 '16 at 21:21

HINT:

Note that we have an indeterminate form. So, apply L'Hospital's Rule and the Fundamental Theorem of Calculus.

• Use of L'Hospital is unnecessary here. – Paramanand Singh Jun 23 '16 at 9:11
• @ParamanandSingh Yes, it is unnecessary. One can simply use the MVT for integrals instead. – Mark Viola Jun 23 '16 at 13:09