Question

lim(x→π/2)([1 - tan(x/2)][1 - sinx])/([1 + tan(x/2)][π - 2x]^3 is

Answer 1

Consider the given expression,

$limit_{x\rightarrow \frac{\Pi }{2}}\frac{(1-\tan \frac{x}{2})(1-\sin x)}{(1+\tan\frac{x}{2})(\Pi-2x^{3})}$

Expressing tan in the form of sin and cos, we get

$=limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})(1-\sin x)}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\Pi-2x^{3})}$

Multiplying and dividing by $(\cos \frac{x}{2}-\sin \frac{x}{2})$, we get

= $limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})(1-\sin x)}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})(\Pi-2x^{3})}$

= $limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})^{2}(1-\sin x)}{(\cos^{2} \frac{x}{2}-\sin^{2} \frac{x}{2})(\Pi-2x^{3})}$

= $limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos^{2} \frac{x}{2}+\sin^{2} \frac{x}{2}-2\sin\frac{x}{2}\cos \frac{x}{2})(1-\sin x)}{(\cos x)(\Pi-2x^{3})}$

=$limit_{x\rightarrow \frac{\Pi }{2}}\frac{(1-\sin x)(1-\sin x)}{(\cos x)(\Pi-2x^{3})}$

= $limit_{x\rightarrow \frac{\Pi }{2}}\frac{(1-\sin x)^{2}}{(\cos x)(\Pi-2x^{3})}$

Let $\Pi-2x = t$

When x tends to $\frac{\Pi}{2}$ , t tends to 0.

Also x = $\frac{\Pi }{2}-\frac{t}{2}$

$=limit_{t\rightarrow 0}\frac{(1-\sin (\frac{\Pi }{2}-\frac{t}{2}))^{2}}{(\cos (\frac{\Pi }{2}-\frac{t}{2}))(\Pi-2(\frac{\Pi }{2}-\frac{t}{2})^{3})}$

= $limit_{t\rightarrow 0}\frac{(1-\cos \frac{t}{2})^{2}}{(\sin \frac{t}{2})({t}^{3})}$

= $limit_{t\rightarrow 0}\frac{(2\sin^{2}\frac{t}{4})^{2}}{(2\sin \frac{t}{4}\cos\frac{t}{4})({t}^{3})}$

= $limit_{t\rightarrow 0}\frac{(2\sin^{3}\frac{t}{4})}{(\cos\frac{t}{4})({t}^{3})}$

$=limit_{t\rightarrow 0}\frac{(2\sin^{3}\frac{t}{4})}{t^{3}}$

= $2 \times \frac{1}{64}$

= $\frac{1}{32}$

So, the final answer is $\frac{1}{32}$.

Answer 2

Answer:

Step-by-step explanation:

limit_{x\rightarrow \frac{\Pi }{2}}\frac{(1-\tan \frac{x}{2})(1-\sin x)}{(1+\tan\frac{x}{2})(\Pi-2x^{3})}  

Expressing tan in the form of sin and cos, we get

=limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})(1-\sin x)}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\Pi-2x^{3})}

Multiplying and dividing by  (\cos \frac{x}{2}-\sin \frac{x}{2}) , we get

=  limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})(1-\sin x)}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})(\Pi-2x^{3})}  

=  limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})^{2}(1-\sin x)}{(\cos^{2} \frac{x}{2}-\sin^{2} \frac{x}{2})(\Pi-2x^{3})}  

=  limit_{x\rightarrow \frac{\Pi }{2}}\frac{(\cos^{2} \frac{x}{2}+\sin^{2} \frac{x}{2}-2\sin\frac{x}{2}\cos \frac{x}{2})(1-\sin x)}{(\cos x)(\Pi-2x^{3})}  

= limit_{x\rightarrow \frac{\Pi }{2}}\frac{(1-\sin x)(1-\sin x)}{(\cos x)(\Pi-2x^{3})}  

=  limit_{x\rightarrow \frac{\Pi }{2}}\frac{(1-\sin x)^{2}}{(\cos x)(\Pi-2x^{3})}  

Let  \Pi-2x = t  

When x tends to  \frac{\Pi}{2}  , t tends to 0.

Also x =   \frac{\Pi }{2}-\frac{t}{2}  

=limit_{t\rightarrow 0}\frac{(1-\sin (\frac{\Pi }{2}-\frac{t}{2}))^{2}}{(\cos (\frac{\Pi }{2}-\frac{t}{2}))(\Pi-2(\frac{\Pi }{2}-\frac{t}{2})^{3})}  

=  limit_{t\rightarrow 0}\frac{(1-\cos \frac{t}{2})^{2}}{(\sin \frac{t}{2})({t}^{3})}  

=  limit_{t\rightarrow 0}\frac{(2\sin^{2}\frac{t}{4})^{2}}{(2\sin \frac{t}{4}\cos\frac{t}{4})({t}^{3})}  

=  limit_{t\rightarrow 0}\frac{(2\sin^{3}\frac{t}{4})}{(\cos\frac{t}{4})({t}^{3})}  

=limit_{t\rightarrow 0}\frac{(2\sin^{3}\frac{t}{4})}{t^{3}}  

=  2 \times \frac{1}{64}  

=  \frac{1}{32}