Question

If \(\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin x^{2} d x}{x^{n}}\) is a non zero definite number, then value of \(\mathrm{n}\) is (A) 1 (B) 3 (C) 5 (D) 4

Short answer

Based on the step-by-step solution, the correct answer is (A) 1 as the limit is non-zero and definite when \(n=2\). The given limit expression is: \(\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin x^{2} d x}{x^{n}}\). After evaluating the integral and applying L'Hôpital's rule, we got the simplified limit expression as \(\lim _{x \rightarrow 0} \frac{-\sin x^2}{n x^{n-2}}\). Analyzing the limit, we concluded that the non-zero definite value of the limit occurs at \(n=2\).

Step-by-step solution

Evaluate the given integral

Use substitution method to evaluate integral \(\int_{0}^{x^{2}} \sin x^{2} d x\). Let \(y = x^2\), then we get \(dy = 2x\, d x\). The limits transform from \(0 \to x^2\) to \(0 \to y\). The integral now becomes: $$\frac{1}{2}\int_{0}^{y} \sin y\, dy$$ Now, evaluate the integral: $$\frac{1}{2}\int_{0}^{y} \sin y\,dy = -\frac{1}{2}\left[ \cos y\right]_{0}^{y} = \frac{1}{2}\left(\cos 0 - \cos y \right) = \frac{1}{2}\left(1 - \cos y \right)$$ Now, replace \(y\) by \(x^2\): $$\frac{1}{2}\left(1 - \cos x^2 \right)$$

Simplify the expression

Now, we insert the result of the integral into the given limit expression: $$\lim _{x \rightarrow 0} \frac{\frac{1}{2}\left(1 - \cos x^{2} \right)}{x^{n}}$$

Apply L'Hopital's rule

To evaluate the limit, we will use L'Hôpital's rule. It states that if the limit is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then, $$\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim _{x \rightarrow 0} \frac{f'(x)}{g'(x)}$$ provided both derivatives exist and are continuous in a neighborhood of \(0\). So, we need to find the derivative of both the numerator and denominator with respect to \(x\). We have: $$f(x) = \frac{1}{2}\left(1 - \cos x^{2} \right)$$ $$g(x) = x^n$$ Find their derivatives with respect to \(x\): $$f'(x) = -x\sin x^2$$ $$g'(x) = nx^{n-1}$$ Apply L'Hôpital's rule: $$\lim _{x \rightarrow 0} \frac{-x\sin x^2}{nx^{n-1}} = \lim _{x \rightarrow 0} \frac{-\sin x^2}{n x^{n-2}}$$

Analyze the limit

Observe that if \(n > 2\), then the limit becomes zero because the denominator goes to infinity. So, we must search for non-zero definite value of the limit in the cases \(n = 1\) and \(n=2\). If \(n=1\), then the limit becomes \(\lim_{x\to 0} \frac{-\sin x^2}{x}\), which is also a zero limit because the numerator goes to zero while the denominator remains constant. So we are left with \(n=2\). Now, evaluate the limit at \(n=2\): $$\lim _{x \rightarrow 0} \frac{-\sin x^2}{2x} = \lim _{x \rightarrow 0} -\frac{1}{2}\cot x^{2}$$ This limit is non-zero and definite. Hence, the value of \(n = 2\). The correct answer is (A) 1.