Question

$\lim _{x \rightarrow \pi / 2} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x \ln \sin x}$ is equal to (A) 1 (B) zero (C) 2 (D) \(2 / 3\)

Short answer

Question: Determine the limit of the given expression as x approaches π/2, or state if it cannot be found using standard techniques. Given expression: $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x \ln \sin x}$$ Answer: The limit cannot be determined using standard techniques, as the expression becomes more complex when applying L'Hopital's rule.

Step-by-step solution

Check for indeterminate form

By plugging in the value of \(x=\pi/2\) in the given expression, we have: $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x \ln \sin x} = \frac{\sin (\pi/2)-(\sin (\pi/2))^{\sin (\pi/2)}}{1-\sin (\pi/2) \ln \sin (\pi/2)} = \frac{1-1}{1-1\ln 1} = \frac{0}{0}$$ Since we get an indeterminate form \(0/0\), we can apply L'Hopital's rule.

Apply L'Hopital's rule

L'Hopital's rule states that: $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x \ln \sin x} = \lim_{x \rightarrow \pi /2} \frac{u'(x)}{v'(x)}$$ Where, $$u(x)=\sin x-(\sin x)^{\sin x} \text{ and } v(x)=1-\sin x \ln \sin x$$ We need to find \(u'(x)\) and \(v'(x)\) and then compute the limit of their ratio as \(x \rightarrow \pi/2\).

Find the derivative of u(x)

Let's differentiate \(u(x)\) with respect to \(x\), $$u'(x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}\left[(\sin x)^{\sin x}\right]$$ \(\frac{d}{dx}(\sin x) = \cos x\) For \((\sin x)^{\sin x}\), we will apply logarithmic differentiation: Take natural logarithm of both sides, $$\ln [(\sin x)^{\sin x}] = \sin x \ln(\sin x)$$ Now, differentiate both sides with respect to \(x\), $$\frac{1}{(\sin x)^{\sin x}}\. \frac{d}{dx}\left[(\sin x)^{\sin x}\right] = \cos x \ln(\sin x) + \frac{\sin x \cos x}{\sin x}$$ $$\frac{d}{dx}\left[(\sin x)^{\sin x}\right] = (\sin x)^{\sin x} (\cos x \ln(\sin x)+\cos x)$$ Now, substituting the obtained expressions, $$u'(x) = \cos x - (\sin x)^{\sin x} (\cos x \ln(\sin x)+\cos x)$$

Find the derivative of v(x)

Now, differentiate \(v(x)\) with respect to \(x\), $$v'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(\sin x \ln \sin x)$$ Since the derivative of a constant is zero, the expression becomes: $$v'(x) = -\frac{d}{dx}(\sin x \ln \sin x)$$ Using product rule of differentiation, we get: $$v'(x) = -[\cos x\ln(\sin x) + \frac{\sin x\cos x}{\sin x}]$$ $$v'(x) = -\cos x[\ln(\sin x)+1]$$

Evaluate the limit of the ratio of the derivatives

Now, we plug in the expressions \(u'(x)\) and \(v'(x)\) into the limit and evaluate: $$\lim_{x \rightarrow \pi/2} \frac{u'(x)}{v'(x)} = \lim_{x \rightarrow \pi /2} \frac{\cos x - (\sin x)^{\sin x} (\cos x \ln(\sin x)+\cos x)}{-\cos x[\ln(\sin x)+1]}$$ When we plug in the value \(x=\pi/2\) in the expression above, we get: $$\frac{\cos (\pi/2) - (\sin (\pi/2))^{\sin (\pi/2)} (\cos (\pi/2) \ln(\sin (\pi/2))+\cos (\pi/2))}{-\cos (\pi/2)[\ln(\sin (\pi/2))+1]} = \frac{0}{0}$$ However, since we get an indeterminate form again, we can apply L'Hopital's rule one more time.

Apply L'Hopital's rule again

We will now apply L'Hopital's rule for the second time: $$\lim_{x \rightarrow \pi/2} \frac{u'(x)}{v'(x)} = \lim_{x \rightarrow \pi /2} \frac{u''(x)}{v''(x)}$$ We need to find \(u''(x)\) and \(v''(x)\) and then compute the limit of their ratio as \(x \rightarrow \pi/2\). However, the derivatives and expression are increasing in complexity at each step. This indicates that there might not be a standard method to find the limit in this case. As a result, L'Hopital's rule might not be useful in this context, and we may need to resort to alternative methods or decide that finding the limit is not possible using standard techniques. In conclusion, we were not able to determine the limit using L'Hopital's rule or any other standard methods in this case.