Question

The function f(x) = $(sin 2x)^{tan^2 2x}$ is not defined at x = π/4. The value of f(π/4) so that f is continuous at x = π/4, is

• A. $\sqrt{e}$
• B. 1/$\sqrt{e}$
• C. 2
• D. None of these

Answer 1

Answer: (B)

1/$\sqrt{e}$

Answer 2

We need to find

$$f\left(\frac{\pi}{4}\right) = \lim_{x\to \pi/4} (\sin2x)^{\tan^2 2x}.$$

Taking the natural logarithm, let

$$y = (\sin2x)^{\tan^2 2x} \quad \Rightarrow \quad \ln y = \tan^2 2x \cdot \ln(\sin2x).$$

As $x \to \frac{\pi}{4}$, we have $2x \to \frac{\pi}{2}$. Set

$$u = \frac{\pi}{2} - 2x, \quad \text{so that } u \to 0.$$

Now, express the functions in terms of $u$:

• $\tan2x = \tan\left(\frac{\pi}{2} - u\right) = \cot u \approx \frac{1}{u}$,
• $\sin2x = \sin\left(\frac{\pi}{2} - u\right) = \cos u \approx 1 - \frac{u^2}{2}$,
• Hence, $\ln(\sin2x) \approx \ln\left(1-\frac{u^2}{2}\right) \approx -\frac{u^2}{2}$.

So,

$$\tan^2 2x \cdot \ln(\sin2x) \approx \frac{1}{u^2} \left(-\frac{u^2}{2}\right) = -\frac{1}{2}.$$

Thus, taking the limit,

$$\ln y \to -\frac{1}{2} \quad \Rightarrow \quad y \to e^{-1/2} = \frac{1}{\sqrt{e}}.$$

Core Explanation

Let $y=(\sin 2x)^{\tan^2 2x}$. Taking $\ln y$ gives $\tan^2(2x) \ln(\sin2x)$. Near $x=\pi/4$, set $u=\pi/2-2x$: then $\tan2x\approx1/u$ and $\ln(\sin2x)\approx -u^2/2$, so product $\approx -1/2$. Thus, $y\to e^{-1/2}=\frac{1}{\sqrt{e}}$.