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What is $x \rightarrow \frac{\pi}{2} f(x)$ equal to ?
Consider the function $f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}$
Where $x \neq \frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=\lambda$
Options:
Consider the function $f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}$
Where $x \neq \frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=\lambda$
Solution:
1183 Upvotes
Verified Answer
The correct answer is:
$1 / 8$
$\lim _{x \rightarrow \frac{\pi}{2}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2} \frac{1-\sin \mathrm{x}}{(\pi-2 \mathrm{x})^{2}}}$
$=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{-\cos \mathrm{x}}{(\pi-2 \mathrm{x})(-2)}$
$=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{\cos \mathrm{x}}{4(\pi-2 \mathrm{x})}$
and $=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{-\sin \mathrm{x}}{4(-2)}=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{\sin \mathrm{x}}{8}$
$=\frac{1}{8} \cdot \sin \frac{\pi}{2}=\frac{1}{8} \times 1=\frac{1}{8}$
$=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{-\cos \mathrm{x}}{(\pi-2 \mathrm{x})(-2)}$
$=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{\cos \mathrm{x}}{4(\pi-2 \mathrm{x})}$
and $=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{-\sin \mathrm{x}}{4(-2)}=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{\sin \mathrm{x}}{8}$
$=\frac{1}{8} \cdot \sin \frac{\pi}{2}=\frac{1}{8} \times 1=\frac{1}{8}$
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