limx→π4(csc2x−2tan2x)(cot⁡x−1) without L'Hôpitals Rule.

Hint: (csc2x−2tan2x)(cot⁡x−1)=csc2−2−2tan2x+2cot⁡x−1 =csc2x−2cot⁡x−1+2tan⁡x1−tan2x1−tan⁡x =cot2x−1cot⁡x−1+2tan⁡x1−tan2x1−tan⁡x =cot⁡x+1+2tan⁡x(1+tan⁡x) It is easier to take the limit now

This limit can actually be calculated quite directly. You only need tan⁡π4=1 Rewrite the expression as follows and set t=tan⁡x and consider t→1: (csc2x−2tan2x)(cot⁡x−1)=cos2xsin2x+1−2tan2x1tan⁡x−1 =t=tan⁡x1t2+1−2t21t−1 =1+t2−2t4t(1−t) =(1−t)(1+t)(1+2t2)t(1−t) =(1+t)(1+2t2)t→t→16

I use (at the end) cos⁡x−sin⁡x=2((12)cos⁡x−(12)sin⁡x)=2(sin⁡(π4)cos⁡x−cos⁡(π4)sin⁡x)=2sin⁡(π4−x)v=limx→π4csc2⁡x−2tan2⁡xcot⁡x−1=limx→π41sin2⁡x−2sin2⁡xcos2⁡xcos⁡x

Hint: (csc2x−2tan2x)(cot⁡x−1)=csc2−2−2tan2x+2cot⁡x−1 =csc2x−2cot⁡x−1+2tan⁡x1−tan2x1−tan⁡x =cot2x−1cot⁡x−1+2tan⁡x1−tan2x1−tan⁡x =cot⁡x+1+2tan⁡x(1+tan⁡x) It is easier to take the limit now

This limit can actually be calculated quite directly. You only need tan⁡π4=1 Rewrite the expression as follows and set t=tan⁡x and consider t→1: (csc2x−2tan2x)(cot⁡x−1)=cos2xsin2x+1−2tan2x1tan⁡x−1 =t=tan⁡x1t2+1−2t21t−1 =1+t2−2t4t(1−t) =(1−t)(1+t)(1+2t2)t(1−t) =(1+t)(1+2t2)t→t→16

I use (at the end) cos⁡x−sin⁡x=2((12)cos⁡x−(12)sin⁡x)=2(sin⁡(π4)cos⁡x−cos⁡(π4)sin⁡x)=2sin⁡(π4−x)v=limx→π4csc2⁡x−2tan2⁡xcot⁡x−1=limx→π41sin2⁡x−2sin2⁡xcos2⁡xcos⁡x

We have an answer to "limx→π4(csc2x−2tan2x)(cot⁡x−1) without L'Hôpitals Rule."

High School
Geometry
Trigonometry
Trigonometric equations and identities

lugreget9

Answered question

2021-12-30

lim_{x \to \frac{π}{4}} \frac{(\csc^{2} x - 2 \tan^{2} x)}{(\cot x - 1)}

without L'Hôpitals Rule.

Answer & Explanation

Jenny Bolton

Beginner2021-12-31Added 32 answers

Hint:

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

= \cot x + 1 + 2 \tan x (1 + \tan x)

It is easier to take the limit now

Serita Dewitt

Beginner2022-01-01Added 41 answers

This limit can actually be calculated quite directly. You only need

\tan \frac{π}{4} = 1

Rewrite the expression as follows and set

t = \tan x

and consider

t \to 1

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

\overset{t = \tan x}{=} \frac{\frac{1}{t^{2}} + 1 - 2 t^{2}}{\frac{1}{t} - 1}

= \frac{1 + t^{2} - 2 t^{4}}{t (1 - t)}

= \frac{(1 - t) (1 + t) (1 + 2 t^{2})}{t (1 - t)}

= \frac{(1 + t) (1 + 2 t^{2})}{t} \overset{t \to 1}{\to} 6

Vasquez

Expert2022-01-09Added 669 answers

I use (at the end)
$\begin{matrix} \cos x - \sin x = \sqrt{2} ((\frac{1}{\sqrt{2}}) \cos x - (\frac{1}{\sqrt{2}}) \sin x) \\ = \sqrt{2} (\sin (\frac{π}{4}) \cos x - \cos (\frac{π}{4}) \sin x) \\ = \sqrt{2} \sin (\frac{π}{4} - x) \\ v = lim_{x \to \frac{π}{4}} \frac{\csc^{2} x - 2 \tan^{2} x}{\cot x - 1} \\ = lim_{x \to \frac{π}{4}} \frac{\frac{1}{\sin^{2} x} - 2 \frac{\sin^{2} x}{\cos^{2} x}}{\cos x} \end{matrix}$

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