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$\lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}=$
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Verified Answer
The correct answer is:
$\sqrt{2}$
$\begin{aligned}
& \lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-\pi / 4} \\
& =\lim _{\alpha \rightarrow \pi / 4}\left\{\frac{\sqrt{2}\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}-\cos \alpha \cdot \frac{1}{\sqrt{2}}\right)}{\left(\alpha-\frac{\pi}{4}\right)}\right\} \\
& =\sqrt{2} \lim _{\alpha \rightarrow \pi / 4} \frac{\sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2} \times 1=\sqrt{2} .
\end{aligned}$
Aliter : Apply L-Hospital's rule,
$\lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-(\pi / 4)}=\lim _{\alpha \rightarrow \pi / 4} \frac{\cos \alpha+\sin \alpha}{1}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} .$
& \lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-\pi / 4} \\
& =\lim _{\alpha \rightarrow \pi / 4}\left\{\frac{\sqrt{2}\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}-\cos \alpha \cdot \frac{1}{\sqrt{2}}\right)}{\left(\alpha-\frac{\pi}{4}\right)}\right\} \\
& =\sqrt{2} \lim _{\alpha \rightarrow \pi / 4} \frac{\sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2} \times 1=\sqrt{2} .
\end{aligned}$
Aliter : Apply L-Hospital's rule,
$\lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-(\pi / 4)}=\lim _{\alpha \rightarrow \pi / 4} \frac{\cos \alpha+\sin \alpha}{1}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} .$
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