Ayaan Barr

2022-07-16

Evaluate: $\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\theta -\frac{\pi }{4}}$

Kroatujon3

Expert

Add and subtract $\mathrm{cos}\left(\frac{\pi }{4}\right)=\mathrm{sin}\left(\frac{\pi }{4}\right)$ in the numerator to get:
$\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\theta -\frac{\pi }{4}}=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{cos}\left(\frac{\pi }{4}\right)-\left(\mathrm{sin}\theta -\mathrm{sin}\left(\frac{\pi }{4}\right)\right)}{\theta -\frac{\pi }{4}}$
$=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{cos}\left(\frac{\pi }{4}\right)}{\theta -\frac{\pi }{4}}-\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{sin}\theta -\mathrm{sin}\left(\frac{\pi }{4}\right)}{\theta -\frac{\pi }{4}}=-\mathrm{sin}\left(\frac{\pi }{4}\right)-\mathrm{cos}\left(\frac{\pi }{4}\right)=-\sqrt{2}$
where we used the definition for derivatives of sine and cosine.

Gauge Terrell

Expert

Let $\theta -\frac{\pi }{4}=x$. Hence, $\theta =\frac{\pi }{4}+x$ and
Hence,
$\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\theta -\frac{\pi }{4}}=\underset{x\to 0}{lim}\frac{\mathrm{cos}\left(\frac{\pi }{4}+x\right)-\mathrm{sin}\left(\frac{\pi }{4}+x\right)}{x}=-\sqrt{2}\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=-\sqrt{2}.$

Do you have a similar question?