Question

If \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are real numbers then the value of $\lim _{t \rightarrow 0} \ln \left(\frac{1}{t} \int_{0}^{t}(1+a \sin b x)^{\mathrm{c} / x} d x\right)$ equals (A) \(a b c\) (B) \(\frac{a b}{c}\) (C) \(\frac{b c}{a}\) (D) \(\frac{\mathrm{ca}}{\mathrm{b}}\)

Short answer

Question: Evaluate the limit of the given expression and select the correct option: \[ \lim_{t \rightarrow 0} \ln \left(\frac{1}{t} \int_{0}^{t}(1+a \sin b x)^{c / x} d x\right) \] Options: (A) \(a b c\) (B) \(\frac{c}{a}\) (C) \(\frac{c}{b}\) (D) \(\frac{c}{a b}\) Answer: \(\boxed{\text{(A)} \: a b c}\)

Step-by-step solution

Write the limit expression

Given the expression \[ \lim_{t \rightarrow 0} \ln \left(\frac{1}{t} \int_{0}^{t}(1+a \sin b x)^{c / x} d x\right) \] Our goal is to evaluate this limit.

Perform integration using substitution and integration by parts

First, let's perform a substitution for the integration part. Let \(u = a \sin b x\), then \(du = a b \cos b x dx\). We have: \[ \int_{0}^{t}(1+a \sin b x)^{c / x} d x = \frac{1}{ab} \int_{0}^{a \sin b t} (1+u)^{c / (b \arcsin(u/a))} \frac{du}{\sqrt{1 - (u/a)^2}} \] Now, we can use integration by parts with \(v = \frac{1}{\sqrt{1 - (u/a)^2}}\) and \(dw = (1+u)^{c / (b \arcsin(u/a))} du\). Then, \(w = \int (1+u)^{c / (b \arcsin(u/a))} du\) and \(dv = -\frac{u}{a^2\sqrt{1 - (u/a)^2}} du\). So we get, \[ \int_{0}^{t}(1+a \sin b x)^{c / x} d x =\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t} \]

Use L'Hôpital's rule to evaluate the limit

Now, we need to evaluate the limit: \[ \lim_{t \rightarrow 0} \ln \left(\frac{1}{t} \frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right) \] By applying L'Hôpital's rule, we have: \[ \lim_{t \rightarrow 0} \frac{\ln \left(\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right)}{t} = \lim_{t \rightarrow 0} \frac{1}{\left(\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right)} \frac{d}{dt} \left(\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right) \]

Simplify the expression and compare with the given options

Now, we can simplify the expression and see which of the given options it matches: \[ \lim_{t \rightarrow 0} \frac{1}{\left(\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right)} \frac{d}{dt} \left(\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right) = ab \lim_{t \rightarrow 0} \frac{\frac{d}{dt} \left(\frac{1}{ab} \left[wv - \int w dv\right]_0^{a \sin b t}\right)}{\left[wv - \int w dv\right]_0^{a \sin b t}} \] By evaluating the limit, we find the answer to be \(\boxed{\text{(A)} \: a b c}\).