# Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$

Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.

My Attempt: \begin{align} \lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}} &=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}} \\ &=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}. \end{align}

How do I proceed?

• L'Hôpital's rule en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule – ChargeShivers Jul 19 '17 at 1:40
• @ChargeShivers This question is now tagged with limits-without-lhopital – nalzok Jul 19 '17 at 2:07
• you can use taylor series – Marios Gretsas Jul 19 '17 at 2:20
• Taylor series and L'Hopital are basically the same thing, just written differently. – DanielV Jul 19 '17 at 2:26
• You can also use sin((u-pi/4)+pi/4), and cos((u-pi/4)+pi/4). Let z=u-pi/4, then apply sum of sine and cosine property, and expand. Ultimately, you will then need to evaluate limit of sin(z)/z as it approaches zero which is 1. – tighten Jul 19 '17 at 2:36

Let $\theta-\frac{\pi}{4}=x$. Hence, $\theta=\frac{\pi}{4}+x$ and

Hence, $$\lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}=\lim_{x\rightarrow0}\frac{\cos\left(\frac{\pi}{4}+x\right)-\sin\left(\frac{\pi}{4}+x\right)}{x}=-\sqrt2\lim_{x\rightarrow0}\frac{\sin{x}}{x}=-\sqrt2.$$

Add and subtract $\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)$ in the numerator to get:

$$\lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}} = \lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \cos\left(\frac{\pi}{4}\right) - \left(\sin \theta - \sin\left(\frac{\pi}{4}\right)\right)}{\theta - \dfrac {\pi}{4}}$$ $$= \lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \cos\left(\frac{\pi}{4}\right)}{\theta - \dfrac {\pi}{4}} - \lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\sin \theta - \sin\left(\frac{\pi}{4}\right)}{\theta - \dfrac {\pi}{4}} = -\sin\left(\dfrac {\pi}{4}\right) - \cos\left(\dfrac {\pi}{4}\right) = -\sqrt{2}$$

where we used the definition for derivatives of sine and cosine.

• Nice. My only change would be to remove the limit signs until the very end; they clog up the works. – zhw. Jul 19 '17 at 2:32

$$\lim_{\theta \to \frac {\pi}{4}} \frac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$$ Let $x:=\theta-\pi /4$, the limit you want to find equals to $$\lim_{x \to 0} \frac {\cos (x+\pi /4) - \sin (x+\pi /4)}{x}$$ $$=\lim_{x \to 0} -\frac {\sqrt 2(\sin (x+\pi /4)\cos (\pi /4)-\sin (\pi /4)\cos (x+\pi /4))}{x}$$ As something we (intentionally) got in the bracket is the compound angle formula for $\sin$, we have the above limit to equal $$\lim_{x \to 0} -\frac {\sqrt 2(\sin (x+\pi /4-\pi /4))}{x}$$ $$=\lim_{x \to 0} -\frac {\sqrt 2\sin{x}}{x}$$ $$=-\sqrt 2$$

Remember that $\displaystyle \cos(\alpha) - \cos(\beta) = 2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\beta-\alpha}{2}\right)$. Hence we obtain \begin{align*} \cos(\theta) - \sin(\theta) = \cos(\theta) - \cos\left(\frac{\pi}{2}-\theta\right) = 2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}-\theta\right) = \sqrt{2}\sin\left(\frac{\pi}{4}-\theta\right) \end{align*} Finally, we get: \begin{align*} \lim_{\theta\rightarrow\pi/4}\frac{\cos(\theta)-\sin(\theta)}{\theta-\displaystyle\frac{\pi}{4}} = \lim_{\theta\rightarrow\pi/4}\frac{\sqrt{2}\displaystyle\sin\left(\frac{\pi}{4}-\theta\right)}{\theta-\displaystyle\frac{\pi}{4}} = \lim_{x\rightarrow 0}\frac{-\sqrt{2}\sin(x)}{x} = -\sqrt{2} \end{align*} Where it has been used the well known result $$\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$$

Note: $cos(π/4) = sin(π/4) = 1/√2$.

$\frac{cos(\theta) - sin(\theta)}{\theta - π/4} =$

√2 $\frac{sin(π/4) cos(\theta) - cos(π/4)sin(\theta)}{\theta - π/4} =$

√2$\frac{sin(π/4 - \theta)}{\theta - π/4} =$

-√2$\frac{sin(x)}{x}$ , where x: = $\theta - π/4$.

$\lim_{x \rightarrow 0}$ (-√2 $\frac{sin(x)}{x}) =$ -√2.

Both numerator and denominator go to zero, so use L'hopital's. Take derivative of numerator and denominator. Get $(-\sin\theta-\cos\theta)/1$. Take $\lim_\theta\rightarrow \frac {\pi}{4}$. Get $(-\sqrt {2}/2-\sqrt {2}/2)/1=-\sqrt 2$.