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 $\lim\limits_{x\to 0}\frac{1}{x}$.

  • 1
    $\begingroup$ Note that the splitting would have worked fine if the derivative $F'(0)\neq 0$. $\endgroup$ – Paramanand Singh Mar 12 '20 at 0:57

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