Question

The limit
\(\lim\limits_{a\rightarrow0}\left(\frac{\int\limits_{0}^{a}\sin(x^2)dx}{\int\limits_{0}^{a}(\ln(x+1))^2dx}\right)\)
is

• A. 0
• B. 1
• C. \(\frac{\pi}{e}\)
• D. non-existent

Answer 1

The Correct Option is B

To solve the limit:

\(\lim\limits_{a\rightarrow0}\left(\frac{\int\limits_{0}^{a}\sin(x^2)dx}{\int\limits_{0}^{a}(\ln(x+1))^2dx}\right)\), we need to employ L'Hôpital's rule, which is applicable when evaluating limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

1. First, check the forms of the numerator and denominator as \(a \rightarrow 0\):

   • The numerator: \(\int\limits_{0}^{a}\sin(x^2)dx\) approaches 0 as \(a \rightarrow 0\) because the integral's range shrinks to a point.
   • The denominator: \(\int\limits_{0}^{a}(\ln(x+1))^2dx\) also approaches 0 as \(a \rightarrow 0\) by the same reasoning.

   Thus, the original limit is indeed of the indeterminate form \(\frac{0}{0}\).

2. Apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator with respect to \(a\):

   • Differentiating the numerator: \(\frac{d}{da}\left[\int\limits_0^a \sin(x^2) \, dx\right] = \sin(a^2)\).
   • Differentiating the denominator: \(\frac{d}{da}\left[\int\limits_0^a (\ln(x+1))^2 \, dx\right] = (\ln(a+1))^2\).

   The limit thus transforms to:

   \(\lim\limits_{a\rightarrow0}\frac{\sin(a^2)}{(\ln(a+1))^2}\)

3. Evaluate the new limit as \(a \rightarrow 0\):

   As \(a \rightarrow 0\), \(\sin(a^2) \sim a^2\) and \((\ln(a+1))^2 \sim a^2\) because:

   • \(\sin(a^2) \approx a^2\) by the Taylor expansion of \(\sin(x)\) around 0.
   • \(\ln(a+1) \approx a\), so \((\ln(a+1))^2 \approx a^2\).

   Therefore, the limit simplfies to:

   \(\lim\limits_{a\rightarrow0}\frac{a^2}{a^2} = 1\)

4. Conclude that the limit is:

   The final result of the limit is 1. Therefore, the correct answer is 1.

In summary, using L'Hôpital's Rule and basic approximations of functions near zero, we have shown that the given limit evaluates to 1.