4
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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*}

$\endgroup$
5
$\begingroup$

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 $h\to0$

$\endgroup$
  • $\begingroup$ @user80551, how about this? $\endgroup$ – lab bhattacharjee Dec 15 '13 at 10:14
  • $\begingroup$ This one is better $\endgroup$ – user80551 Dec 15 '13 at 10:16
  • 1
    $\begingroup$ @user80551, actually the angles in the other answer being fraction, obscures the clarity of the answer $\endgroup$ – lab bhattacharjee Dec 15 '13 at 10:17
3
$\begingroup$

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

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

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

Can you please consolidate ?

$\endgroup$
  • $\begingroup$ Step 3, how did you get $\frac \pi 4 - x$ in the denominator? $\endgroup$ – user80551 Dec 15 '13 at 9:56
  • $\begingroup$ @user80551, I have divided the denominator & the numerator by $\frac\pi4-x$ which is $\ne0$ as $x\to\frac\pi4$ $\endgroup$ – lab bhattacharjee Dec 15 '13 at 10:01
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.