How to find : $\lim_{x \to 1/4}\frac{1-\tan(\pi x)}{2x-\sqrt{x}}$

Question

How to find :

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

---

My Try :

$$x-\frac{1}{4}=t \to x=t+\frac{1}{4}$$

And:

$$\tan\pi(t+\frac{1}{4})=\tan(\pi t +\frac{\pi}{4})$$

So we have :

$$\lim_{t \to 0}\frac{1-\tan(\pi t +\frac{\pi}{4})}{2(t+\frac{1}{4})-\sqrt{t+\frac{1}{4}}}$$

but now what ?

Answer 1

$$\lim_{t \to 0}\frac{1-\tan(\pi t +\frac{\pi}{4})}{2(t+\frac{1}{4})-\sqrt{t+\frac{1}{4}}}=\lim_{t \to 0}\frac{1-\frac{\tan\pi t+1}{1-\tan\pi t}}{\left(2\sqrt{t+\frac{1}{4}}-1\right)\sqrt{t+\frac{1}{4}}}=$$ $$=\lim_{t\rightarrow0}\frac{-4\tan\pi t}{2\sqrt{t+\frac{1}{4}}-1}=\lim_{t\rightarrow0}\frac{-4\sin\pi t\left(2\sqrt{t+\frac{1}{4}}+1\right)}{4t}=-2\pi\lim_{t\rightarrow0}\frac{\sin\pi t}{\pi t}=-2\pi.$$

Answer 2

\begin{align} \lim_{x\to\frac14}\dfrac{1-\tan\pi x}{2x-\sqrt{x}} &= \lim_{x\to\frac14}\dfrac{\tan\dfrac{\pi}{4}-\tan\pi x}{2x-\sqrt{x}}\times\dfrac{2x+\sqrt{x}}{2x+\sqrt{x}} \\ &= \lim_{x\to\frac14}\frac{\sin\left(\frac{\pi}{4}-\pi x\right)}{\cos\frac{\pi}{4}\cos\pi x}\dfrac{2x+\sqrt{x}}{4x^2-x} \\ &= \lim_{x\to\frac14}\frac{\sin\pi\left(\frac{1}{4}-x\right)}{\cos\frac{\pi}{4}\cos\pi x}\dfrac{2x+\sqrt{x}}{-4x(\frac14-x)} \\ &= \lim_{x\to\frac14}\dfrac{\sin\pi\left(\frac{1}{4}-x\right)}{\pi(\frac14-x)} \dfrac{\pi}{-4x\cos\dfrac{\pi}{4}\cos\pi x} (2x+\sqrt{x}) \\ &= \dfrac{\pi}{-\dfrac{\sqrt{2}}{2}\dfrac{\sqrt{2}}{2}} (\frac12+\frac12) \\ &= \color{blue}{-2\pi} \end{align}

Answer 3

If you know about Taylor expansions, this is a practical solution.

Built aroud $x=\frac 14$, you have $$\tan(\pi x)=1+2 \pi \left(x-\frac{1}{4}\right)+2 \pi ^2 \left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right)$$ $$\sqrt x=\frac{1}{2}+\left(x-\frac{1}{4}\right)-\left(x-\frac{1}{4}\right)^2+O\left(\left(x- \frac{1}{4}\right)^3\right)$$ All of the above make $$\frac{1-\tan(\pi x)}{2x-\sqrt{x}}=\frac{-2 \pi \left(x-\frac{1}{4}\right)-2 \pi ^2 \left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right) } { \left(x-\frac{1}{4}\right)+\left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right)}$$ $$\frac{1-\tan(\pi x)}{2x-\sqrt{x}}=\frac{-2 \pi -2 \pi ^2 \left(x-\frac{1}{4}\right)+O\left(\left(x-\frac{1}{4}\right)^2\right) } { 1+\left(x-\frac{1}{4}\right)+O\left(\left(x-\frac{1}{4}\right)^2\right)}$$ which shws the limit.

If you continue with long division, you should get $$\frac{1-\tan(\pi x)}{2x-\sqrt{x}}=-2 \pi -2\pi\left( \pi -1\right) \left(x-\frac{1}{4}\right)+O\left(\left(x-\frac{1}{4}\right)^2\right)$$ which shows how is approached the limit.

Answer 4

Well, if $g(x)=\tan(\pi x)$, then $g'(x)=\pi\sec^2(\pi x)$, and $g'(1/4)=\pi(\sqrt2)^2=2\pi$. But from the definition of the derivative, $$ g'(1/4)=\lim_{x\to1/4}\frac{\tan(\pi x)-1}{x-\frac14}=-4\lim_{x\to1/4}\frac{1-\tan(\pi x)}{4x-1}\,, $$ so that $$ \lim_{x\to1/4}\frac{1-\tan(\pi x)}{4x-1}=-\pi/2\,. $$ On the other hand, \begin{align} \frac{1-\tan(\pi x)}{2x-\sqrt x}&=\frac{\bigl(1-\tan(\pi x)\bigr)(2x+\sqrt x\,)}{4x^2-x}\\ &=\frac{1-\tan(\pi x)}{4x-1}\cdot\frac{2x+\sqrt x}{x}\,, \end{align} in which product the left-hand factor has limit $-\pi/2$ at $1/4$, while the right-hand factor has value at $1/4$ equal to $4$. Thus the desired limit is $-2\pi$.