Trigonometric limit $\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

Question

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?

Answer 1

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 $$\lim_{x\to\pi/4}\frac{1-1+2(x-\pi/4)}{1-\sqrt2\left[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}(x-\pi/4)\right]}=\lim_{x\to\pi/4}\frac{2(x-\pi/4)}{(x-\pi/4)}=2$$

Answer 2

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 $2$.

Answer 3

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*}

Answer 4

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!

Answer 5

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.