Evaluation of trignometric limit

Question

I want to find the following limit without L'Hospital.

$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$

Maybe I should try to get rid of the radical.

Answer 1

As the OP suggested openness to asymptotic analysis, we proceed accordingly.

First, we note the expansion of the numerator is given by

$$\begin{align} 1+\tan^{1/3} x&=1+\left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\ &=\frac23 (x-3\pi/4)+O(x-3\pi/4)^2 \tag 1 \end{align}$$

whereas the expansion of the denominator is given by and

$$\begin{align} 1-2\cos^2x&=-\cos 2x\\\\ &=-2(x-3\pi/4)+O(x-3\pi/4)^3 \tag 2 \end{align}$$

Thus, putting $(1)$ and $(2)$ together yields

$$\begin{align} \frac{1+\tan^{1/3} x}{1-2\cos^2x}&=\frac{\frac23 (x-3\pi/4)+O(x-3\pi/4)^2}{-2(x-3\pi/4)+O(x-3\pi/4)^3}\\\\ &=-\frac13+O(x-3\pi/4)\\\\ &\to -\frac13\,\,\text{as}\,\,x\to 3\pi/4 \end{align}$$

Therefore, we have that the limit of interest is

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 3\pi/4}\frac{1+\tan^{1/3} x}{1-2\cos^2x}=-\frac13}$$

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NOTES:

To arrive at $(1)$, we first note that for $f(x)=\tan x$, $f'(x)=\sec^2 x$ and $f''(x)=2\sec^2 x\tan x$.

Therefore, $f(3\pi/4)=-1$ and $f'(3\pi/4)=\frac12$ so that

$$\tan x=-1+2(x-3\pi/4)+O(x-3\pi/4)^2$$

Then, using the expansion for $(-1+x)^{1/3}=-1+\frac13x+O(x^2)$, we see that

$$\begin{align} \tan^{1/3} x&= \left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\ &=-1+\frac23(x-3\pi/4)+O(x-3\pi/4)^2 \end{align}$$

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To arrive at $(2)$, we first note the trigonometric identity $\cos 2x=2\cos^2-1$.

Then, for $g(x)=1-2\cos^2 x=-\cos 2x$, we have $g'(x)=2\sin 2x$, $g''(x)=4\cos 2x$, and $g'''(x)=-8 \sin 2x$.

Therefore, $g(3\pi/4)=0$ and $g'(3\pi/4)=-2$, and $g''(3\pi/4)=0$ so that

$$-\cos 2x=-2(x-3\pi/4)+O(x-3\pi/4)^3$$

Answer 2

We will use the fact that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This will help us rationalize the numerator. With $a=1$and $b=(\tan x)^{1/3}$ we can write \begin{align*} \frac{1+(\tan x)^{1/3}}{1-2\cos^2x}& = \frac{1+(\tan x)^{1/3}}{1-2\cos^2x} \cdot \left[\color{blue}{\frac{1-(\tan x)^{1/3}+(\tan x)^{2/3}}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}}\right]\\ & = \frac{1+\tan x}{1-2\cos^2x}\cdot \left[\frac{1}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}\right] \end{align*} Now use the following identity: $$\cos 2x=2 \cos^2 x -1 =\frac{1-\tan^2x}{1+\tan^2x},$$ to get \begin{align*} \frac{1+(\tan x)^{1/3}}{1-2\cos^2x}&= \frac{1+\tan^2 x}{\tan x-1}\cdot \left[\frac{1}{1-(\tan x)^{1/3}+(\tan x)^{2/3}}\right] \end{align*} Now use the fact that $x \to 3\pi/4$ to get the limit as $-1/3$.

Answer 3

let us make a change of variable $x = 3\pi/4 + h.$ then we have $$\begin{align}\tan(x) &= \tan(3\pi/4 + h) = \frac{-1 + \tan h}{1 + \tan h} \\ &= (\sin h - \cos h)(\sin h + \cos h \cdots)^{-1} \\ &=(-1 + h \cdots)(1+h +\cdots)^{-1} \\ &=(-1 + h \cdots)(1-h +\cdots) \\ &=-1+2h+\cdots\\ \tan^{1/3} x &=-1 + \frac23 h+\cdots\\ 1+\tan^{1/3} x &= \frac23 h+\cdots \tag 1\\ \sqrt 2\cos x &= \sqrt 2\cos(3\pi/4+h)= -\cos h - \sin h\\ &=-1 - h + h^2/2+\cdots\\ 2\cos^2 x &= 1+2h+\cdots\\ 1-2\cos^2 x &= -2h+\cdots \tag 2 \end{align}$$

from $(1)$ and $(2),$ $$\lim_{x \to 3\pi/4} \frac{1+\tan^{1/3} x}{1-2\cos^2 x }=\frac{2h/3+\cdots}{-2h + \cdots} = -\frac 13. $$