deiteresfp

2021-12-28

Evaluate: $\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\theta -\frac{\pi }{4}}$
My Attempt:
This limit takes $\frac{0}{0}$ form when $\theta =\frac{\pi }{4}$. Here, $\frac{0}{0}$ form is an indeterminate form. So how do I make it determinate?

nghodlokl

Without lHospital:

vicki331g8

Since your limit is in the form $\frac{00}{}$, you can use LHopitals rule.
$\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\theta -\frac{\pi }{4}}=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\frac{d}{d\theta }\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)}{\frac{d}{d\theta }\left(\theta -\frac{\pi }{4}\right)}=\underset{\theta \to \frac{\pi }{4}}{lim}\left(-\mathrm{sin}\theta -\mathrm{cos}\theta \right)=\dots$
Now that your limit is not indeterminate, you can evaluate your limit by substitution.

nick1337

If you know that for small x you have $\mathrm{sin}x\approx x$ (from Taylor expansion or l'Hopital), you can write $\theta -\frac{\pi }{4}=x$. Then you can show
$\mathrm{cos}\left(x+\frac{\pi }{4}\right)-\mathrm{sin}\left(x+\frac{\pi }{4}\right)=-\sqrt{2}\mathrm{sin}x$
Then your limit is $-\sqrt{2}$

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