# Evaluating $\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

How can I evaluate $$\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ without L'Hopital rule. Using L'Hopital rule, it evaluates to 2. Is there a way to do it without using L'Hopital?

Multiply by the conjugate and use trig identities, factoring appropriately: \begin{align*} \lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x} &= \lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x} \cdot \frac{1 + \sqrt{2}\sin x}{1 + \sqrt{2}\sin x} \\ &= \lim_{x\to\frac{\pi}{4}}\frac{(1-\tan x)(1 + \sqrt{2}\sin x)}{1 - 2\sin^2 x} \\ &= \lim_{x\to\frac{\pi}{4}}\frac{(1-\frac{\sin x}{\cos x})(1 + \sqrt{2}\sin x)}{(1 - \sin^2 x) - \sin^2 x} \\ &= \lim_{x\to\frac{\pi}{4}}\frac{(1-\frac{\sin x}{\cos x})(1 + \sqrt{2}\sin x)}{\cos^2 x - \sin^2 x} \cdot \frac{\cos x}{\cos x} \\ &= \lim_{x\to\frac{\pi}{4}}\frac{(\cos x - \sin x)(1 + \sqrt{2}\sin x)}{\cos x(\cos x - \sin x)(\cos x + \sin x)} \\ &= \lim_{x\to\frac{\pi}{4}}\frac{1 + \sqrt{2}\sin x}{\cos x(\cos x + \sin x)} \\ &= \frac{1 + \sqrt{2}\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}(\cos \frac{\pi}{4} + \sin \frac{\pi}{4})} \\ &= \frac{1 + \sqrt{2}(\frac{1}{\sqrt 2})}{\frac{1}{\sqrt 2}(\frac{1}{\sqrt 2} + \frac{1}{\sqrt 2})} = \frac{1 + 1}{\frac{1}{\sqrt 2}(\frac{2}{\sqrt 2})} = \frac{2}{2/2} = 2 \end{align*}

Setting $\displaystyle\frac\pi4-x=2h$

$\displaystyle\tan x=\tan\left(\frac\pi4-2h\right)=\frac{1-\tan2h}{1+\tan2h}$ $\displaystyle\implies 1-\tan x=\frac{2\tan2h}{1+\tan2h}$

and $\displaystyle\sin x=\sin\left(\frac\pi4-2h\right)=\frac{\cos2h-\sin2h}{\sqrt2}$

$$\lim_{x\to\frac{\pi}4}\frac{1-\tan x}{1-\sqrt{2}\sin x} =\lim_{h\to0}\frac{2\tan2h}{(1+\tan2h)(1-\cos2h+\sin2h)}$$

$$=\lim_{h\to0}\frac{\sin2h}{1-\cos2h+\sin2h}\cdot\lim_{h\to0}\frac2{\cos2h(1+\tan2h)}$$

Now the second limit is easy,

For the first using Double-Angle Formulas, $\displaystyle\frac{\sin2h}{1-\cos2h+\sin2h}=\frac{2\sin h\cos h}{1-(1-2\sin^2h)+2\sin h\cos h}$ $\displaystyle=\frac{\cos h}{\sin h+\cos h}$ if $\sin h\ne0$ which is true as $h\to0$

• This one is better – user80551 Dec 15 '13 at 10:16
• @user80551, actually the angles in the other answer being fraction, obscures the clarity of the answer – lab bhattacharjee Dec 15 '13 at 10:17

$$\lim_{x\to\frac{\pi}4}\frac{1-\tan x}{1-\sqrt{2}\sin x} =\frac1{\sqrt2}\lim_{x\to\frac{\pi}4}\frac{\tan\frac\pi4-\tan x}{\sin\frac\pi4-\sin x}$$

$$=\frac1{\sqrt2}\lim_{x\to\frac{\pi}4}\frac{\sin\left(\frac\pi4-x\right)}{\cos\frac\pi4\cos x}\frac1{\left(\sin\frac\pi4-\sin x\right)}$$

Now, $\displaystyle\lim_{x\to\frac{\pi}4}\frac{\sin\left(\frac\pi4-x\right)}{\frac\pi4-x}=1$

Again using Prosthaphaeresis Formula, $\displaystyle\sin\frac\pi4-\sin x=2\sin\frac{\frac\pi4-x}2\cos\frac{\frac\pi4+x}2$

Now, $\displaystyle\lim_{x\to\frac{\pi}4}\frac{\sin\left(\frac{\frac\pi4-x}2\right)}{\frac{\frac\pi4-x}2}=1$

• Step 3, how did you get $\frac \pi 4 - x$ in the denominator? – user80551 Dec 15 '13 at 9:56
• @user80551, I have divided the denominator & the numerator by $\frac\pi4-x$ which is $\ne0$ as $x\to\frac\pi4$ – lab bhattacharjee Dec 15 '13 at 10:01
Another way to do it is to use Taylor expansions of $\tan(x)$ and $\sin(x)$ built around $x = \frac \pi 4$. Using only the very first terms, they write
$$\tan(x) = 1+2 \left(x-\frac{\pi }{4}\right)+2 \left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$
$$\sin(x) = \frac{1}{\sqrt{2}}+\frac{x-\frac{\pi }{4}}{\sqrt{2}}-\frac{\left(x-\frac{\pi }{4}\right)^2}{2 \sqrt{2}}+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ Replacing in the expression
$$\frac{1-\tan x}{1-\sqrt{2}\sin x}=\frac{-2 \left(x-\frac{\pi }{4}\right)-2 \left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right) } {-\left(x-\frac{\pi }{4}\right)+\frac{1}{2} \left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right) }$$
Now, the limit is obvious. You can have more performing the long division and get $$\frac{1-\tan x}{1-\sqrt{2}\sin x}=2+3 \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ which shows the limit and how it is approached.