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The limit is $$\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ I was able to solve it using L'hopital and the answer that I got was $2$.

Can you please confirm if the answer is right and suggest some other way to evaluate the limit without using L'hopital?

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Multiply both deminator and numerator to the $\left( 1+\sqrt { 2 } \sin { x } \right) $ and $\left( 1+\tan { x } \right) $ respectively \begin{align*} \lim_{x\to\frac{\pi}{4}}{\frac{1-\tan x}{1-\sqrt{2}\sin x}}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\left(1-\tan^2\!x\right)\left(1+\sqrt2\sin x\right)}{\left(1-2\sin^2\!x\right)\left(1+\tan x\right)}}={} \\ {}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\frac{\cos(2x)}{\cos^2\!x}\left(1+\sqrt2\sin x\right)}{\cos(2x)\left(1+\tan x\right)}}={} \\ {}={}&\lim_{x\to\frac{\pi}{4}}\frac{\left(1+\sqrt2\sin x\right)}{\cos^2\!x\left(1+\tan x\right)}=2. \end{align*}

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  • $\begingroup$ can you see your edits $\endgroup$ – Archis Welankar Apr 17 '16 at 16:11
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Let $x=y+\frac{\pi}4$. Then $$\tan x=\frac{\tan y+1}{1-\tan y}$$ and $$\sin x=\frac1{\sqrt2}\sin y+\frac1{\sqrt2}\cos y$$ $$\begin{align}\frac{1-\tan x}{1-\sqrt2\sin x}&=\frac{1-\frac{\tan y+1}{1-\tan y}}{1-\sin y-\cos y}\\ &=\frac{-2\tan y}{(1-\tan y)(1-\sin y-\cos y)}\\ &=\frac{-2\sin y}{(\cos y-\sin y)(1-\sin y-(1-2\sin^2(\frac y2)))}\\ &=\frac2{(\cos y-\sin y)(1-\frac{2\sin^2(\frac y2)}{2\sin(\frac y2)\cos(\frac y2)})}\\ &=\frac2{(\cos y-\sin y)(1-\tan(\frac y2))}\end{align}$$ Take limit as $y\rightarrow0$ to get $2$.

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You can use Taylor expansions. The Taylor expansion of a function $f(x)$ around a point $x=a$ can be written as the infinite sum $$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+...$$ You can find more information about Taylor expansion in https://es.wikipedia.org/wiki/Serie_de_Taylor.

If you expand the trigonometric functions around $x=\pi/4$ up to first order you find $$\tan(x)=1+\left.1/\cos(x)^2\right|_{x=\pi/4}(x-\pi/4)+...=1+2(x-\pi/4)+...\\ \sin(x)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}(x-\pi/4)+...$$ Then, considering the terms in the polynomial up to first order $(x-\pi/4)$, the limit can be written as $$\lim_{x\to\pi/4}\frac{1-1+2(x-\pi/4)}{1-\sqrt2\left[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}(x-\pi/4)\right]}=\lim_{x\to\pi/4}\frac{2(x-\pi/4)}{(x-\pi/4)}=2$$

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  • $\begingroup$ Seoanes, Can you please post the Taylor expansion of sin and tan I don't know them $\endgroup$ – Gem Apr 17 '16 at 16:08
  • $\begingroup$ I have added some more information about the way to compute the Taylor expansion. In this case it is sufficient to know the expansion up to first order (you don't need to determine all the polynomial orders to solve your problem). $\endgroup$ – seoanes Apr 17 '16 at 16:18
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Here we proceed without using L'Hospital's Rule or other methodologies based on differential calculus, but use only basic trigonometric identities. To that end, we write

$$\begin{align} \frac{1-\tan(x)}{1-\sqrt{2}\sin(x)}&=\frac{-\sqrt{2}\sin(x-\pi/4)}{\cos(x)(1-\sin(x-\pi/4)-\cos(x-\pi/4))}\\\\ &=-\frac{2}{\sqrt{2}\cos(x)}\,\frac{2\sin\left(x-\pi/4\right)}{2\sin^2\left(\frac{x-\pi/4}{2}\right)-\sin(x-\pi/4)}\\\\ &=-\frac{2}{\sqrt{2}\cos(x)}\,\frac{\cos\left(\frac{x-\pi/4}{2}\right)}{\sin\left(\frac{x-\pi/4}{2}\right)-\cos\left(\frac{x-\pi/4}{2}\right)}\\\\ \end{align}$$

Finally, letting $x\to \pi/4$ we find the coveted limit to be

$$\lim_{x\to \pi/4}\frac{1-\tan(x)}{1-\sqrt{2}\sin(x)}=2$$

as expected!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola May 9 '16 at 16:53
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Let $f(x) = 1-\tan x, g(x) = 1-\sqrt 2\sin x.$ The expression equals

$$\frac{f(x) - f(\pi/4)}{g(x) - g(\pi/4)} = \frac{(f(x) - f(\pi/4))/(x-\pi/4)}{g(x) - g(\pi/4)/(x-\pi/4)} .$$

By definition of the derivative, this $\to f'(\pi/4)/g'(\pi/4)$ as $x\to \pi/4.$ This quotient of derivatives is easy to compute.

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