Question

$\operatorname{Limit}_{x \rightarrow \pi / 2} \frac{\left(1-\tan \frac{x}{2}\right)(1-\sin x)}{\left(1+\tan \frac{x}{2}\right)(\pi-2 x)^{3}}$ is

• A. 1/16
• B. -1/16
• C. 1/32
• D. -1/32

Answer 1

Answer: (C)

1/32

Answer 2

To evaluate the limit $\operatorname{Limit}_{x \rightarrow \pi / 2} \frac{\left(1-\tan \frac{x}{2}\right)(1-\sin x)}{\left(1+\tan \frac{x}{2}\right)(\pi-2 x)^{3}}$, we observe that as $x \rightarrow \pi/2$, the expression takes the indeterminate form $0/0$.

We use the substitution method. Let $x = \frac{\pi}{2} - h$.
As $x \rightarrow \frac{\pi}{2}$, $h \rightarrow 0$.

Now, substitute $x = \frac{\pi}{2} - h$ into the expression:

1. Numerator terms:

   • $1 - \tan \frac{x}{2} = 1 - \tan \left(\frac{\pi}{4} - \frac{h}{2}\right)$
     Using the tangent subtraction formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
     $\tan \left(\frac{\pi}{4} - \frac{h}{2}\right) = \frac{\tan(\pi/4) - \tan(h/2)}{1 + \tan(\pi/4)\tan(h/2)} = \frac{1 - \tan(h/2)}{1 + \tan(h/2)}$
     So, $1 - \tan \frac{x}{2} = 1 - \frac{1 - \tan(h/2)}{1 + \tan(h/2)} = \frac{(1 + \tan(h/2)) - (1 - \tan(h/2))}{1 + \tan(h/2)} = \frac{2\tan(h/2)}{1 + \tan(h/2)}$.

   • $1 - \sin x = 1 - \sin \left(\frac{\pi}{2} - h\right)$
     Using the identity $\sin(\pi/2 - \theta) = \cos \theta$:
     $1 - \sin x = 1 - \cos h$.

2. Denominator terms:

   • $1 + \tan \frac{x}{2} = 1 + \tan \left(\frac{\pi}{4} - \frac{h}{2}\right)$
     $1 + \tan \frac{x}{2} = 1 + \frac{1 - \tan(h/2)}{1 + \tan(h/2)} = \frac{(1 + \tan(h/2)) + (1 - \tan(h/2))}{1 + \tan(h/2)} = \frac{2}{1 + \tan(h/2)}$.

   • $\pi - 2x = \pi - 2\left(\frac{\pi}{2} - h\right) = \pi - \pi + 2h = 2h$.
     Therefore, $(\pi - 2x)^3 = (2h)^3 = 8h^3$.

Now substitute these back into the limit expression: $\operatorname{Limit}_{h \rightarrow 0} \frac{\left(\frac{2\tan(h/2)}{1 + \tan(h/2)}\right)(1-\cos h)}{\left(\frac{2}{1 + \tan(h/2)}\right)(8h^3)}$ Cancel out the common term $\frac{2}{1 + \tan(h/2)}$ from the numerator and denominator: $\operatorname{Limit}_{h \rightarrow 0} \frac{\tan(h/2)(1-\cos h)}{8h^3}$ We can rewrite this expression to use standard limits: $\operatorname{Limit}_{h \rightarrow 0} \frac{\tan(h/2)}{h} \cdot \frac{1-\cos h}{h^2} \cdot \frac{1}{8}$ We know the following standard limits:

• $\operatorname{Limit}_{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1$
• $\operatorname{Limit}_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$

Apply these to our expression:

• For $\frac{\tan(h/2)}{h}$: Multiply and divide by 2 to match the form $\frac{\tan \theta}{\theta}$:
  $\operatorname{Limit}_{h \rightarrow 0} \frac{\tan(h/2)}{h} = \operatorname{Limit}_{h \rightarrow 0} \frac{\tan(h/2)}{h/2} \cdot \frac{1}{2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$.

• For $\frac{1-\cos h}{h^2}$: This is a direct application of the standard limit:
  $\operatorname{Limit}_{h \rightarrow 0} \frac{1-\cos h}{h^2} = \frac{1}{2}$.

Now, substitute these limit values back into the expression: $\left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{8}\right) = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}$

The value of the limit is $\frac{1}{32}$.