MHB Limit of Ratio of Difference of Trig Functions at $\pi/4$

[Petrus]

$$\lim_{x \to \pi/4} \frac{1-\tan(x)}{\sin(x)-\cos(x)}$$
progress:
I start with rewriting $\tan(x)=\frac{\sin(x)}{\cos(x)}$

 

[chisigma]

Re: Trig limit

$$\lim_{x \to \pi/4} \frac{1-\tan(x)}{\sin(x)-\cos(x)}$$
progress:
I start with rewriting $\tan(x)=\frac{\sin(x)}{\cos(x)}$

Try writing...

$\displaystyle \frac{1 - \tan x}{\sin x - \cos x} = - \frac{1}{\cos x}\ \frac{1- \frac{\sin x}{\cos x}} {1 - \frac{\sin x}{\cos x}} = - \frac{1}{\cos x}$ (1)

Kind regards

$\chi$ $\sigma$

 

[Moriarty1]

Re: Trig limit

Let $\displaystyle \ell = \lim_{x \to \frac{\pi}{4}}\frac{1-\tan{x}}{\cos{x}-\sin{x}}.$ Sub $x = \tan^{-1}{t} $ then $ x \to \frac{\pi}{4} \implies t \to 1$; also $\tan{x} = t$, $\displaystyle \cos{x} = \frac{1}{\sqrt{1+t^2}}$ and $\displaystyle \sin{x} = \frac{t}{\sqrt{1+t^2}}$. Thus $\displaystyle \ell = \lim_{t \to 1}\frac{1-t}{\frac{1-t}{\sqrt{1+t^2}}} = \lim_{t \to 1}\sqrt{1+t^2} = \sqrt{2}$.

 

[soroban]

Hello, Petrus!

A slightly different approach . . .

\lim_{x\to \frac{\pi}{4}} \frac{1-\tan x}{\sin x - \cos x}

\lim_{x\to\frac{\pi}{4}} \frac{1 - \frac{\sin x}{\cos x}}{\sin x - \cos x} \;=\; \lim_{x\to\frac{\pi}{4}}\frac{\frac{\cos x - \sin x}{\cos x}}{\sin x - \cos x} \;=\;\lim_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{\cos x(\sin x - \cos x)}

. . . =\;\lim_{x\to\frac{\pi}{4}}\frac{-(\sin x - \cos x)}{\cos x(\sin x - \cos x)} \;=\;\lim_{x\to\frac{\pi}{4}}\frac{-1}{\cos x} \;\;\cdots\;\; \text{etc.}

 

[Prove It]

$$\lim_{x \to \pi/4} \frac{1-\tan(x)}{\sin(x)-\cos(x)}$$
progress:
I start with rewriting $\tan(x)=\frac{\sin(x)}{\cos(x)}$

Since this goes to $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$, L'Hospital's Rule can be used.

$\displaystyle \begin{align*} \lim_{x \to \frac{\pi}{4}}{\frac{1 - \tan{(x)}}{\sin{(x)} - \cos{(x)}}} &= \lim_{x \to \frac{\pi}{4}}{\frac{-\sec^2{(x)}}{\cos{(x)} + \sin{(x)}}} \\ &= \frac{-\left( \sqrt{2} \right)^2}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} \\ &= \frac{-2}{\frac{2}{\sqrt{2}}} \\ &= -\sqrt{2} \end{align*}$