# GRE math question: $\lim_{x \to 0} \left[ \frac{1}{x^2} \int_0^x \frac{t + t^2}{1 + \sin t}\, \mathrm{d} t \right]$

This question is from the Princeton Review book Cracking the GRE Mathematics Subject Test, chapter 2, question 7. The question asks to find the following limit:

$$\lim_{x \to 0} \left[ \dfrac{1}{x^2} \int_0^x \dfrac{t + t^2}{1 + \sin t}\, \mathrm{d} t \right]$$

My solution was as follows: let $$F(t)$$ be some antiderivative of $$(t + t^2)/(1 + \sin t)$$. Then, the limit can be written

\begin{align} \lim_{x \to 0} \left[ \dfrac{1}{x^2} (F(x) - F(0)) \right] &= \lim_{x \to 0} \left[ \dfrac{1}{x} \cdot \dfrac{F(x) - F(0)}{x} \right] \\ &= \lim_{x \to 0} \left[ \dfrac{1}{x} \cdot F'(0) \right] = 0 \end{align}

However, the correct answer is $$\dfrac{1}{2}$$, as given here:

Since the integral equals $$0$$ when $$x = 0$$, the limit is of the indeterminate form $$\dfrac{0}{0}$$, so we apply L'Hôpital's rule

$$\lim_{x \to 0}\frac{\int_0^x \dfrac{t + t^2}{1 + \sin t} \, dt}{x^2} = \lim_{x \to 0}\frac{\dfrac{x + x^2}{1 + \sin x}}{2x}$$

$$= \lim_{x \to 0}\frac{x(1 + x)}{2x(1 + \sin x)} = \lim_{x \to 0}\frac{1 + x}{2(1 + \sin x)} = \frac{1}{2}$$

I understand the provided solution, but cannot see why my solution is incorrect?

$$F'(0)$$ does not equal $$\dfrac{F(x)-F(0)}{x}$$, but
$$\lim_{x\to 0}\dfrac{F(x)-F(0)}{x}$$
So (to use your way) you would have to split the limits, but then you get an undefined limit $$\lim\limits_{x\to 0}\frac{1}{x}$$.
• Note that the splitting would have worked fine if the derivative $F'(0)\neq 0$. – Paramanand Singh Mar 12 '20 at 0:57