$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule.

The limit,

$$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$$

is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $$\frac{0}{0}$$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.

Hint:\begin{align*}\frac{\csc^2 x-2\tan^2 x}{\cot x-1}&=\frac{\csc^2 x-2-2\tan^2 x+2}{\cot x-1} ]\\&=\frac{\csc^2 x-2}{\cot x-1}+2\tan x\frac{1-\tan^2 x}{1-\tan x}\\&=\frac{\cot^2 x-1}{\cot x-1}+2\tan x\frac{1-\tan^2 x}{1-\tan x}\\&=\boxed{\cot x+1+2\tan x(1+\tan x)}\end{align*}

It is easier to take the limit now.....

• Much easier than my answer. Dec 4 '20 at 3:39
• @martycohen yours is also nice(+1),although lentgthy it requires a lot of thinking to comup with what you have done Dec 4 '20 at 3:43
• I just plug ahead naively and see what happens, Sometimes it works, sometimes it doesn't. This time it worked. Dec 4 '20 at 3:53
• @martycohen i see, anyway your solution on second reading only seems lengthy maybe bevause you have elaborated the steps very detailed unlike mine. Dec 4 '20 at 3:55
• Very nice. I didn’t think of the “+0” strategy. Thanks! Dec 4 '20 at 4:13

This limit can actually be calculated quite directly. You only need $$\tan \frac{\pi}{4}= 1$$

Rewrite the expression as follows and set $$t=\tan x$$ and consider $$t\to 1:$$

$$\begin{eqnarray*}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)} & = & \frac{\frac{\cos^2 x}{\sin^2 x}+1-2\tan^2x}{\frac 1{\tan x}- 1} \\ & \stackrel{t=\tan x}{=} & \frac{\frac 1{t^2}+1-2t^2}{\frac 1t-1}\\ & = & \frac{1+t^2 - 2t^4}{t(1-t)}\\ & = & \frac{(1-t)(1+t)(1+2t^2)}{t(1-t)}\\ & = & \frac{(1+t)(1+2t^2)}{t} \stackrel{t\to1}{\longrightarrow}6\\ \end{eqnarray*}$$

• Great choice of substitution, it simplifies the limit a lot. Thanks! Dec 4 '20 at 17:16

I use (at the end)

$$\begin{array}\\ \cos x-\sin x &=\sqrt{2}((1/\sqrt{2})\cos x-(1/\sqrt{2})\sin x)\\ &=\sqrt{2}(\sin(\pi/4)\cos x-\cos(\pi/4)\sin x)\\ &=\sqrt{2}\sin(\pi/4-x)\\ \end{array}$$

$$\begin{array}\\ v &=\lim \limits_{x \to \frac\pi4}\dfrac{\csc^2x-2\tan^2x}{\cot x-1}\\ &=\lim \limits_{x \to \frac\pi4}\dfrac{\dfrac1{\sin^2x}-2\dfrac{\sin^2x}{\cos^2x}}{\dfrac{\cos x}{\sin x}-1}\\ &=\lim \limits_{x \to \frac\pi4}\dfrac{\cos^2x-2\sin^4x}{\sin x\cos^3 x-\sin^2x\cos^2x}\\ &=\lim \limits_{x \to \frac\pi4}\dfrac{1-\sin^2x-2\sin^4x}{\sin x\cos x(1-\sin^2x)-\sin^2x(1-\sin^2x)}\\ &=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(1+\sin^2x)}{\sin x(1-\sin^2x)(\cos x-\sin x)}\\ &=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(3/2)}{(\sqrt{2}/2)(\cos x-\sin x)}\\ &=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-2\sin^2x}{(\cos x-\sin x)}\\ &=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}\sin x)}{\cos x-\sin x}\\ &=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}(\sqrt{2}/2))}{\cos x-\sin x}\\ &=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)}{\cos x-\sin x}\\ &=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sqrt{2}\sin(\pi/4-x)}\\ &=3\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sin(\pi/4-x)}\\ &=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}\sin (y+\pi/4)}{\sin(-y)}\\ &=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}(\sin y\cos(\pi/4)+\cos(y)\sin(\pi/4))}{\sin(-y)}\\ &=3\lim \limits_{y \to 0}\dfrac{1-2(\sin y+\cos(y))}{\sin(-y)}\\ &=3\lim \limits_{y \to 0}\dfrac{-2\sin y}{-\sin(y)}\\ &=6\\ \end{array}$$

• Very interesting approach, I like it. I had it exactly like you up to the 5th line, but as the expression kept getting longer I just decided to try something else. Thanks for the answer. Dec 4 '20 at 4:18

$$\frac{\csc^2(x)-2\tan^2(x)}{\cot (x)-1}=2 \tan (x)+\cot (x)+2 \sec ^2(x)-1$$

The denominator $$\cot x-1$$ resembles $$t-1$$ too much to not be taken into consideration.

You surely know that $$\csc^2x=1+\cot^2x,\qquad \tan x=\frac{1}{\cot x}$$ so the numerator becomes, with $$\cot x=t$$, $$1+t^2-\dfrac{2}{t^2}=\dfrac{t^4+t^2-2}{t^2}$$ Never mind if you don't see the factorization $$t^4+t^2-2=(t^2-1)(t^2+2)$$; you do know that a factor $$t-1$$ can be found! With synthetic division you find $$\begin{array}{r|rrrr|r} & 1 & 0 & 1 & 0 & -2 \\ 1 & & 1 & 1 & 2 & 2 \\\hline & 1 & 1 & 2 & 2 & 0\end{array}$$ and thus $$t^4+t^2-2=(t-1)(t^3+t^2+2t+2)$$ and you can rewrite your limit in the form $$\lim_{x\to\pi/4}\frac{\cot^3x+\cot^2x+2\cot x+2}{\cot^2x}=6$$ If you had seen $$t^4+t^2-2=(t^2-1)(t^2+2)=(t-1)(t+1)(t^2+2)$$, you'd get the equivalent form $$\lim_{x\to\pi/4}\frac{(\cot x+1)(\cot^2x+2)}{\cot^2x}$$

• The (t-1) factor is easily seen from (t^2-1). Very intuitive approach, thanks for the detailed answer. Dec 4 '20 at 17:27