Question

$\lim _{x \rightarrow 0^{+}} \int_{x}^{2 x} \frac{\sin ^{\mathrm{m}} \mathrm{t}}{t^{\mathrm{n}}}(\mathrm{m}, \mathrm{n}, \in \mathrm{N})$ equals (A) 0 if \(m \geq n\) (B) \(\ln 2\) if \(\mathrm{n}-\mathrm{m}=1\) (C) \(+\infty\) if \(n-m=1\) (D) None of these

Short answer

Based on the step-by-step solution and analysis above, the short answer to the problem is as follows: The limit of the given integral as \(x\) approaches \(0+\) depends on the relationship between \(m\) and \(n\): - If \(m \geq n\), the limit equals \(0\). - If \(n-m=1\), the limit equals \(\ln 2\). - If \(m < n - 1\), the limit does not match any of the given options, and is considered "None of these." The limit does not reach \(+\infty\) in any case.

Step-by-step solution

Define given limit

First, let's write down the given limit we need to evaluate: $$ \lim_{x \to 0^+} \int_{x}^{2x} \frac{\sin^m t}{t^n} dt. $$

Apply L'Hospital's Rule if needed

We need to check if L'Hospital's Rule could be applicable first, before calculating the limit of the integral. $$ \lim_{x \to 0^+} \frac{\int_{x}^{2x} \sin^m t dt}{\int_x^{2x}t^n dt} $$ But since we have integrals in our expression, we cannot apply L'Hospital's Rule directly.

Calculate integral using substitution

Use the substitution: \(u = \frac{t}{x} \implies t = xu \implies dt = x du\) to change the bounds and simplify the integral: $$ \lim_{x \to 0^+} \int_{1}^{2} \frac{\sin^m (xu)}{(xu)^n} x du = \lim_{x \to 0^+} \int_{1}^{2} x^{1-n} \frac{\sin^m (xu)}{u^n} du. $$

Evaluate the limit

Take the limit of the expression as \(x\) approaches \(0+\): $$ \lim_{x \to 0^+} \int_{1}^{2} x^{1-n} \frac{\sin^m (xu)}{u^n} du $$ We can analyze this expression based on different cases: 1. Case: \(m \geq n\) $$ \lim_{x \to 0^+} x^{1-n} \int_{1}^{2} \frac{\sin^m (xu)}{u^n} du = 0 $$ Since \(\lim_{x \to 0^+} x^{1-n} = 0\) for \(m \geq n\). 2. Case: \(n - m = 1\) $$ \lim_{x \to 0^+} \int_{1}^{2} \frac{\sin^{n-1} (xu)}{u^n} du $$ As \(x \to 0^+\), \(\sin^{n-1} (xu) \to (xu)^{n-1}\), then: $$ \lim_{x \to 0^+} \int_{1}^{2} \frac{(xu)^{n-1}}{u^n} du = \lim_{x \to 0^+} x^{1-n} \int_{1}^{2} \frac{1}{u} du = \ln 2 $$ 3. Case: \(n - m > 1\) or \(m < n - 1\) $$ \lim_{x \to 0^+} x^{1-n} \int_{1}^{2} \frac{\sin^m (xu)}{u^n} du $$ Since \(\lim_{x \to 0^+} x^{1-n} = 0\), this case has a limit of \(0\), but since it doesn't match any of the given cases in the options, it is considered "None of these." After analyzing different cases, we can conclude that: (A) The limit equals 0 if \(m \geq n\) (B) The limit equals \(\ln 2\) if \(n-m=1\) (C) \(+\infty\) is not reached in any case (D) "None of these" corresponds to cases with \(m < n - 1\)