Evaluate $ \displaystyle \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}.$
Solution (1)
$ \displaystyle \ \ \ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\left( {\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}\times \frac{{\frac{{\sqrt{2}}}{2}}}{{\frac{{\sqrt{2}}}{2}}}} \right)$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\frac{{\sqrt{2}}}{2}\sin x-\frac{{\sqrt{2}}}{2}\cos x}}{{\frac{{\sqrt{2}}}{2}\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\frac{{\sqrt{2}}}{2}\sin x-\frac{{\sqrt{2}}}{2}\cos x}}{{\frac{{\sqrt{2}}}{2}\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x\cos \frac{\pi }{4}-\cos x\sin \frac{\pi }{4}}}{{\frac{{\sqrt{2}}}{2}\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\ \sqrt{2}\underset{{\left( {x-\frac{\pi }{4}} \right)\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {x-\frac{\pi }{4}} \right)}}{{\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\sqrt{2}\times 1$
$ \displaystyle =\sqrt{2}$
Solution (2)
Let $ \displaystyle t=x-\frac{\pi }{4}$ and hence $ \displaystyle x=t+\frac{\pi }{4}$
When $ \displaystyle x\to \frac{\pi }{4}$, $ \displaystyle t\to 0$.
$ \displaystyle \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}$
$ \displaystyle = \underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {t+\frac{\pi }{4}} \right)-\cos \left( {t+\frac{\pi }{4}} \right)}}{t}$
$ \displaystyle =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin t\cos \frac{\pi }{4}+\cos t\sin \frac{\pi }{4}-\cos t\cos \frac{\pi }{4}+\sin t\cos \frac{\pi }{4}}}{t}$
$ \displaystyle =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{2\sin t\cos \frac{\pi }{4}}}{t}$
$ \displaystyle =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{2\sin t\left( {\frac{{\sqrt{2}}}{2}} \right)}}{t}$
$ \displaystyle =\sqrt{2}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin t}}{t}$
$ \displaystyle =\sqrt{2}\times 1$
$ \displaystyle =\sqrt{2}$
Solution (1)
$ \displaystyle \ \ \ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\left( {\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}\times \frac{{\frac{{\sqrt{2}}}{2}}}{{\frac{{\sqrt{2}}}{2}}}} \right)$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\frac{{\sqrt{2}}}{2}\sin x-\frac{{\sqrt{2}}}{2}\cos x}}{{\frac{{\sqrt{2}}}{2}\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\frac{{\sqrt{2}}}{2}\sin x-\frac{{\sqrt{2}}}{2}\cos x}}{{\frac{{\sqrt{2}}}{2}\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\ \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x\cos \frac{\pi }{4}-\cos x\sin \frac{\pi }{4}}}{{\frac{{\sqrt{2}}}{2}\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\ \sqrt{2}\underset{{\left( {x-\frac{\pi }{4}} \right)\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {x-\frac{\pi }{4}} \right)}}{{\left( {x-\frac{\pi }{4}} \right)}}$
$ \displaystyle =\sqrt{2}\times 1$
$ \displaystyle =\sqrt{2}$
Solution (2)
Let $ \displaystyle t=x-\frac{\pi }{4}$ and hence $ \displaystyle x=t+\frac{\pi }{4}$
When $ \displaystyle x\to \frac{\pi }{4}$, $ \displaystyle t\to 0$.
$ \displaystyle \underset{{x\to \frac{\pi }{4}}}{\mathop{{\lim }}}\,\frac{{\sin x-\cos x}}{{x-\frac{\pi }{4}}}$
$ \displaystyle = \underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {t+\frac{\pi }{4}} \right)-\cos \left( {t+\frac{\pi }{4}} \right)}}{t}$
$ \displaystyle =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin t\cos \frac{\pi }{4}+\cos t\sin \frac{\pi }{4}-\cos t\cos \frac{\pi }{4}+\sin t\cos \frac{\pi }{4}}}{t}$
$ \displaystyle =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{2\sin t\cos \frac{\pi }{4}}}{t}$
$ \displaystyle =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{2\sin t\left( {\frac{{\sqrt{2}}}{2}} \right)}}{t}$
$ \displaystyle =\sqrt{2}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin t}}{t}$
$ \displaystyle =\sqrt{2}\times 1$
$ \displaystyle =\sqrt{2}$
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